Hi Samsfunhouse. Thanks for the question. After we have subtracted the "n squared" sequence from our original sequence, we are left with 6, 8, 10, 12, 14, ... This sequence that we are left with is just the 2 x table, but it has been "shifted up" by 4 (instead of our 2 x table - 2, 4, 6, 8, 10, ... all of our numbers have increased by 4). Because our numbers are 4 more than the 2 x table, the nth term of this sequence is 2n + 4. We then just add this on to the quadratic part of our sequence, which gives us the overall nth term of n^2 + 2n + 4. I hope that makes sense.
Thank you for explaining step by step. Often times steps are left out because it is assumed that we understand and know the steps that are unsaid and left out. There was one step that drove me crazy and kept me search for an answer until I got to you. Thank you so much. BTW, I was trying to figure out how to get the _n2(squared #'s) to subtract from the original sequence to finally determine the nth term. Thanks again!!
you explained this better than my maths teacher. we spent like 3 lessons on this and i didn't understand. i watched this 9 minute video and i understood better 😭
but what about with n squared? because I'm on a problem that says: "The nth term of a sequence is : n squared + 2n... Work out the first 5 terms in the sequences" I dont understand how to work out n squared
@themasterofslaughter7777 To work out the first 5 terms of a sequence, you need to substitute the numbers 1 to 5 into your nth term. For example: The 1st term would be 1^2 + 2(1) = 3 The 2nd term would be 2^2 + 2(2) = 8 Then do the same for n=3, n=4 and n=5. I hope that makes sense.
Hi Stubby Fox, So, at 5:00, the reason why the sequence of 4, 5, 6, 7, 8, ... has the nth term of "n+3" is because it is 3 places off our 1 x table. If you think about what the 1 x table sequence, it would look like 1, 2, 3, 4, 5, .... Notice that each term in our sequence is 3 numbers higher than the 1 x table, which is where the "+3" comes from. I could have also writte "1n+3", but we don't write 1 in front of any letters. I hope that makes sense 🙂
Hi Humzah, What I mean when I say "shifting down our sequence by 3" is that whatever times table our sequence is in, the numbers in our sequence are just 3 less than that. For example, with the 2nd question, where we have the numbers 1, 5, 9, 13, 17, ..., these numbers are similar to the 4 x table (because they go up by 4 each time), but each number is 3 less than the 4 x table. Another way you can think about it is if you compare the sequence to the 4 x table... Our sequence: 1 5 9 13 17... 4 x table: 4 8 12 16 20... Now, hopefully you can see that each number has been moved down 3 places (or "shifted" as the word I used). I hope that helps!
Hi! out of all the videos, websites, the amount of times I read the book I couldn't find the nth term so thaaank you so much for this! Im quiet getting there though but im just confused, what do you mean by shifting up?
Hello 👋.. thanks for the lovely comment. With reference to what you said about "shifting up", I am just describing the difference between a linear sequence and the times tables. For example, if we have the sequence 4, 7, 10, 13, 16,... I would say that this is similar to the 3 x table, but each number has been "shifted up by 1" - 4 is 1 more than 3(1x3), 7 is 1 more than 6(2x3), 10 is 1 more than 9(3x3), 13 is 1 more than 12(4x3), and 16 is 1 more than 15(5x3). For this sequence, the nth term would be 3n+1 because each number is 1 more than the 3 x table. I hope that makes sense.
Hi Monsoon Moon, You are very welcome. If you write down what you are doing and then I can see where you are going wrong so that I can help. Best wishes
@@BeastModeMaths Tyvm :) it looks a lot like the last example you have in the video. 3, 5, 8, 12, 17 +2, +3, +4, +5 +1, +1, +1 0.5n^2 3 5 8 12 17 0.5 2 4.5 8 12.5 2.5 3 3.5 4 4.5 n + 0.5 0.5n^2 + n + 0.5
@@monsoonmoon7751 You have done nearly everything correct - you have just made a small mistake towards the end. You have calculated the quadratic part of the sequence correctly (0.5n^2). You have also written out the 0.5n^2 sequence correctly and subtracted it from the original sequence correctly. The only mistake you made was calculating the nth term of the sequence "2.5, 3, 3.5, 4, 4.5, ...". Because the difference between each term is 0.5, it has to start with 0.5n. Then, the adjustment that we have to make is to +2 each time. Therefore, the nth term of this sequence is 0.5n + 2. So overall, the nth term of the whole quadratic sequence is 0.5n^2 + 0.5n + 2. I hope that helps - well done though. I really liked your workings out!
@@charikag-vm7bd no problem. I've got this same question a lot, so obviously didn't explain it that well. I'll give you an example... Let's say you had the sequence 5, 7, 9, 11, 13,... - you can think of this sequence as THE SAME as the 2 x table, but each number has been "shifted up" by 3 units - 5 is 3 MORE than 2 (1 x 2) - 7 is 3 MORE than 4 (2 x 2) - 9 is 3 MORE than 6 (3 x 2) - 11 is 3 MORE than 8 (4 x 2) - 13 is 3 MORE than 10 (5 x 2) Hope that clears it up! 😎
Hi TJ. I have a video dedicated to why we half the second difference to give us the coefficient on the n^2 term. Here is the link... th-cam.com/video/9cKBB9VE6Sg/w-d-xo.html&ab_channel=BeastModeMaths
Hi BULL DOG, I could go through the process that I went through in the video, but there is something that stands out... The numbers you have listed are the square numbers. The 1st term is 0 squared, the 2nd term is 1 squared, the 3rd term is 2 squared and the 4th term is 3 squared... Each term is 1 less than the term number and then squared. Therefore, the nth term would be (n-1)^2. I hope that makes sense.
Hello Jane, This is a cubic sequence, because the 3rd difference between terms is the same. Let me explain... The 1st difference between terms is 4, 16, 34, 58, 88, ... Therefore, the 2nd difference between terms is 12, 18, 24, 30, ... Therefore, the 3rd difference between terms is 6, 6, 6, ... Because we now know that it is a cubic sequence, it starts with n^3. To calculate the coefficient on the n^3 term, we divide the 3rd difference by 6. So, 6/6 is 1, so our nth term starts with 1n^3. Now we can subtract the 1n^3 sequence away from our original sequence:- Our sequence: 2, 6, 22, 56, 114, 202 1n^3 sequence: 1, 8, 27, 64, 125, 216 Our results is the following sequence: 1, -2, -5, -8, -11, -14. Now we can calculate the nth term of this sequence and add it on to the cubic part that we just figured out. The nth term of the sequence above is -3n+4. Combining everything together, we get n^3 - 3n + 4. I hope that helps.
So after we have taken the n^3 sequence away from our original sequence, we are left with the following sequence... 1, -2, -5, -8, -11, -14. Now, if we calculate the nth term of this sequence, we know it must start with -3n (because it is going down by 3 each time, i.e. it is similar to our -3 x table). Now we just need to work out the adjustment. Can you see that each number is 4 places above our -3 times table? For example... To go from -3 (1 x -3) to 1, we need to add 4 To go from -6 (2 x -3) to -2, we need to add 4 To go from -9 (3 x -3) to -5, we need to add 4 To go from -12 (4 x -3) to -8, we need to add 4 To go from -15 (5 x -3) to -11, we need to add 4 To go from -18 (6 x -3) to -14, we need to add 4 Just to summarise, the sequence above is the same as our -3 times tables, but each number has been shifted up by 4 places, hence -3n+4. I hope that clears it up.
Hi Hasan Ali. For quadratic sequences, the term-to-term rule is different for each term. That is why we need to calculate the 2nd difference. If the second difference was +2, we could say that the term-to-term rule of this sequence was "add 2 more each time".
@@Jeysiiiiii I'm assuming you are referring to the example at the beginning , and the sequence of 6, 8, 10, 12, 14. Notice that the numbers go up by 2 each time. Therefore, we can say that the sequence is similar to the 2 x table. If we compare it to the 2 x table, each number in our sequence is 4 units higher (e.g. 6 is 4 more than 2, 8 is 4 more than 4, 10 is 4 more than 6, etc.). So each number has been "shifted up" by 4 units. That's where we get the 2n+4 from (4 more than the 2 x table). I hope that helps!!
Hello. The "-3" is part of the 4n-3 which is the nth term of the sequence 1, 5, 9, 13, 17. Let me explain this. The numbers go up by 4 each time, which is where we get 4n from. If the sequence was just 4n then the numbers would go 4, 8, 12, 16, 20 (just the 4 x table). Notice that each number in our sequence is 3 less than the 4n sequence. I hope that makes sense!
Hello Aaron, After we subtracted the quadratic part of our sequence away from our original sequence, we were left with the sequence of numbers 4, 5, 6, 7, 8, ... We then need to calculate what the nth term of this sequence is. This sequence is similar to our 1 x table, but each number is just shifted up by 3 places (for example, 4 is 1x1, plus 3) (5 is 1x2, plus 3) (6 is 1x3, plus 3) (7 is 1x4, plus 3) (8 is 1x5, plus 3) So the nth term of this sequence of numbers is 1n+3. Now, our final nth term is the quadratic part and the linear part put together... Quadratic part = 2n^2 Linear part = n+3 Final nth term = 2n^2 + n +3 I hope that helps - I would recommend going back over calculating the nth term of linear sequences first if this part confused you.
Hello ba na na, So, we first need to identify the 1st difference between terms - this is +3, +5, +7, +9. Now we look at the 2nd difference - this is +2, +2, +2. Because the 2nd difference is the same, we know that it is a quadratic sequence, so our nth term starts with __n^2. We just need to work out the number in front of the n^2 term. To do this, we just half the second difference. So half of 2 is 1. Therefore, our nth term starts with 1n^2 (or we could just say n^2). Now we have found out the quadratic part of our sequence, we need to work out the linear part. Do do this, we just subtract our quadratic sequence (that we have just worked out) from our original sequence... Original sequence: 10, 13, 18, 25, 34 n^2 sequence: 1, 4, 9, 16, 25 Difference: 9, 9, 9, 9, 9 Because the difference between the n^2 sequence and our original sequence is 9, we just need to add that on. So our final nth term is n^2 + 9. I hope all that helps!
Hi Abishan Zi. When we are finding the quadratic part of our sequence, we always half the second difference (for example, in the example I go through, the 2nd difference is 2, therefore half of 2 is 1, so it begins with 1n^2. This is the only part where you half the 2nd difference. The part you are referring to at 2:44 is when calculating the nth term of an arithmetic sequence. If the numbers are 6, 8, 10, 12, 14, then the nth term of this sequence is 2n+4. If you are not quite sure about this part, I would recommend watching a video on how to calculate the nth term of an arithmetic sequence. Many thanks.
Hi sfl, The reason why we do the "shifting" is to make the sequence we are left with look the same as one of our times tables. This might not make much sense, but let me use the first example to help. After we have found the quadratic part of our sequence and we have subtracted it from our original sequence, we are left with the numbers 6, 8, 10, 12, 14. This sequence is the same as our 2 times tables, but it has just been "shifted up" by 4. The only time that we would not need to "shift" would be if the sequence we are left with is exactly the same as one of our times tables (e.g. 2, 4, 6, 8, 10). I hope that helps!
كيف يمكنني ايجاد formula لمتتالية تتكون من ارقام صحيحة موجبة وسالبة ؟ وايظا formula اخرى لمتتالية تتكون من ارقام كسرية موجبة وسالبة ؟ ارجو الاجابة لدي امتحان نهائي غدا 😢
Hi Leonie. So for this sequence the first difference goes +6, +10, +14 then the second difference is +4, +4. As the second difference is the same (+4), we half this number, so the quadratic part of our sequence starts with 2n^2. Now we need to subtract the 2n^2 sequence away from our original sequence to see what is left over. original sequence: 2, 8, 18, 32 2n^2 sequence: 2, 8, 18, 32 It turns out that the 2n^2 sequence is our sequence! We don't have to make any adjustment - so our nth term is just 2n^2 I hope that helps.
@@leonietanke8048 I'm not sure exactly what you are asking. Are you trying to work out what the 66nd term of the sequence is? If so, we can substitute 66 into our nth term... = 2 x 66^2 = 2 x 4356 = 8712.
Hi yungogthathoop. Sorry for the late reply. I've only just seen your comment. For the sequence 7, 12, 17, 22, 27, ... if we compare it with the 5 x table, notice how each number in our sequence is 2 bigger than the 5 x table:- 7 = 5 + 2 12 = 10 + 2 17 = 15 + 2 22 = 20 + 2 ...etc That is where the +2 comes from. I hope that helps!
@@jelmayguardiario8166 I have just calculated the "difference between the differences" and when you get to the fourth difference, it is the same. The 4th difference is -8, which means that this sequence could be a quartic sequence. I would not know how to calculate the nth term of this type of sequence though. Sorry I can't be of any more help.
Hi Leonie, This is a geometric sequence because to go from one term to the next we multiply by the same number. For this sequence, to go from one term to the next, we multiply by 5. We start at 1 and then x5 to get the next term (5) We then x5 again to get the next term (25) We then x5 again to get the next term (125) *I think there is a mistake in the sequence that you gave - the next term should say 625 and then 3125 (I think you have just missed 625) So to calculate our nth term, it will be 1 x 5^(n-1). The power of 5 is always 1 less than our term number. The reason for this is because to get our 2nd term, we multiply by 5 once, to get our 3rd term, we multiply by 5 twice, and so on. I have a video dedicated to geometric sequence, which should help!
In the example you are referring to, the sequence that we are left with (after we have subtracted the quadratic sequence away from our original sequene) is 4, 5, 6, 7, 8. This sequence is the same as our 1 x table, but we have just added 3 to each number in the 1 x table. This is what I mean by being "shifted up by 3 places". I hope that makes sense.
Hi T9 HD. I would say that it doesn't matter too much if it takes a long time to do. The most important thing is doing it correctly so that you end up with the correct answer. Once you get more practice, it will soon become easier. Thanks for watching.
The best explanation on actually solving the question I've found on this topic. Thank you and well done!
Thanks very much! ...and you're welcome!
thank you so much i have a maths test today and i was losing about 4 marks in these q every time thanks for saving my life i owe u one
lol
You're welcome. I'm pleased to hear that it has helped!
He replied in 37 min wow😮
This really made sense and you have explained it better than anyone, keep up the great work and videos so I can learn a lot :)
Thank you RiskTaker. I'm thrilled to hear that you enjoyed the video and that it has helped you! 😀
you taught better than my math teacher she dragged a lot you were like snap and it was awesome
Ahh thanks Aqua gaming! Much appreciated 🤟
What do you mean by "shifting up"? I don't understand how you get the last number.
Hi Samsfunhouse. Thanks for the question.
After we have subtracted the "n squared" sequence from our original sequence, we are left with 6, 8, 10, 12, 14, ...
This sequence that we are left with is just the 2 x table, but it has been "shifted up" by 4 (instead of our 2 x table - 2, 4, 6, 8, 10, ... all of our numbers have increased by 4).
Because our numbers are 4 more than the 2 x table, the nth term of this sequence is 2n + 4.
We then just add this on to the quadratic part of our sequence, which gives us the overall nth term of n^2 + 2n + 4.
I hope that makes sense.
where did the reply from beast math gone?
Thank you for explaining step by step. Often times steps are left out because it is assumed that we understand and know the steps that are unsaid and left out. There was one step that drove me crazy and kept me search for an answer until I got to you. Thank you so much. BTW, I was trying to figure out how to get the _n2(squared #'s) to subtract from the original sequence to finally determine the nth term. Thanks again!!
You're welcome L.LeMond. I know how tricky this particular topic can be so I tried to go slowly and include all of the steps. Thanks for the comment 🙂
I have understood clearly how to solve the problems as a result of excellent explanation.
Thanks Eric - that's really kind of you. I'm glad it helped! 🙂
you explained this better than my maths teacher. we spent like 3 lessons on this and i didn't understand. i watched this 9 minute video and i understood better 😭
Wow - thanks for those kind words! 🙂. I'm pleased to hear that you now have a better understanding on this topic.
😍🤩🤩🤩Thank you so much, I'm preparing for exams and this really helped me
You're very welcome, Milena! Best of luck with the exams 🤞
Thank u soo much sir🙏
I have my IGCSE exam tomorrow!
You're welcome, Hriday Menon. Good luck tomorrow!
So far this is the most helpful vid about sequences , you cleared everything up to me ,thank you so much for that effort 🤍🤍🤍
I'm pleased you foud it helpful, Mona. Thanks for the nice comment :-)
You just Got One More Subscriber Cause you are doing an Awesome Job!
Thanks Study Only. Appreciate your support.
Excellent explanation .
Thank you, Kumar 🙂
i love you 🙏🙏🙏 thank you for helping me revise
You're very welcome! Best of luck with your exams!!
but what about with n squared? because I'm on a problem that says:
"The nth term of a sequence is : n squared + 2n... Work out the first 5 terms in the sequences"
I dont understand how to work out n squared
@themasterofslaughter7777 To work out the first 5 terms of a sequence, you need to substitute the numbers 1 to 5 into your nth term. For example:
The 1st term would be 1^2 + 2(1) = 3
The 2nd term would be 2^2 + 2(2) = 8
Then do the same for n=3, n=4 and n=5.
I hope that makes sense.
Thank you so much it really helped thank you
You're welcome! Thanks.
@@BeastModeMaths your videos are good thanks
why at 5:00 is it put as +3 yet its only going up by 1's
Hi Stubby Fox,
So, at 5:00, the reason why the sequence of 4, 5, 6, 7, 8, ... has the nth term of "n+3" is because it is 3 places off our 1 x table. If you think about what the 1 x table sequence, it would look like 1, 2, 3, 4, 5, .... Notice that each term in our sequence is 3 numbers higher than the 1 x table, which is where the "+3" comes from. I could have also writte "1n+3", but we don't write 1 in front of any letters.
I hope that makes sense 🙂
@@BeastModeMaths thanks
@@Nazeus You're welcome 🙂
Sorry for annoying but what you mean when you say that we are shifting down our sequence by 3
Hi Humzah,
What I mean when I say "shifting down our sequence by 3" is that whatever times table our sequence is in, the numbers in our sequence are just 3 less than that. For example, with the 2nd question, where we have the numbers 1, 5, 9, 13, 17, ..., these numbers are similar to the 4 x table (because they go up by 4 each time), but each number is 3 less than the 4 x table. Another way you can think about it is if you compare the sequence to the 4 x table...
Our sequence: 1 5 9 13 17...
4 x table: 4 8 12 16 20...
Now, hopefully you can see that each number has been moved down 3 places (or "shifted" as the word I used).
I hope that helps!
Hi! out of all the videos, websites, the amount of times I read the book I couldn't find the nth term so thaaank you so much for this! Im quiet getting there though but im just confused, what do you mean by shifting up?
Hello 👋.. thanks for the lovely comment. With reference to what you said about "shifting up", I am just describing the difference between a linear sequence and the times tables. For example, if we have the sequence 4, 7, 10, 13, 16,... I would say that this is similar to the 3 x table, but each number has been "shifted up by 1" - 4 is 1 more than 3(1x3), 7 is 1 more than 6(2x3), 10 is 1 more than 9(3x3), 13 is 1 more than 12(4x3), and 16 is 1 more than 15(5x3). For this sequence, the nth term would be 3n+1 because each number is 1 more than the 3 x table.
I hope that makes sense.
@@BeastModeMaths ah finally get it, thank you thank you so much! :)
@@shylie8207 no problem!
Thanks, well explained. Really Good
Thanks Antony! 🙂
Thanks for the video was really helpful. Still stuck on my sequence though not sure what I am doing wrong.
3, 5, 8, 12, 17...
Hi Monsoon Moon,
You are very welcome. If you write down what you are doing and then I can see where you are going wrong so that I can help.
Best wishes
@@BeastModeMaths Tyvm :) it looks a lot like the last example you have in the video.
3, 5, 8, 12, 17
+2, +3, +4, +5
+1, +1, +1
0.5n^2
3 5 8 12 17
0.5 2 4.5 8 12.5
2.5 3 3.5 4 4.5
n + 0.5
0.5n^2 + n + 0.5
@@monsoonmoon7751 You have done nearly everything correct - you have just made a small mistake towards the end.
You have calculated the quadratic part of the sequence correctly (0.5n^2). You have also written out the 0.5n^2 sequence correctly and subtracted it from the original sequence correctly. The only mistake you made was calculating the nth term of the sequence "2.5, 3, 3.5, 4, 4.5, ...".
Because the difference between each term is 0.5, it has to start with 0.5n. Then, the adjustment that we have to make is to +2 each time. Therefore, the nth term of this sequence is 0.5n + 2.
So overall, the nth term of the whole quadratic sequence is 0.5n^2 + 0.5n + 2.
I hope that helps - well done though. I really liked your workings out!
@@BeastModeMaths okay I see what I did wrong now tyvm :)
hi, tysm for this video!! just a quick q, wdym by shifting up by a number?
@@charikag-vm7bd no problem. I've got this same question a lot, so obviously didn't explain it that well. I'll give you an example... Let's say you had the sequence 5, 7, 9, 11, 13,... - you can think of this sequence as THE SAME as the 2 x table, but each number has been "shifted up" by 3 units
- 5 is 3 MORE than 2 (1 x 2)
- 7 is 3 MORE than 4 (2 x 2)
- 9 is 3 MORE than 6 (3 x 2)
- 11 is 3 MORE than 8 (4 x 2)
- 13 is 3 MORE than 10 (5 x 2)
Hope that clears it up! 😎
@@BeastModeMaths yes it does!! youre the sweetest, god bless ❤❤
Very well explained. Why do you half the coefficient of the n term?
Hi TJ. I have a video dedicated to why we half the second difference to give us the coefficient on the n^2 term. Here is the link... th-cam.com/video/9cKBB9VE6Sg/w-d-xo.html&ab_channel=BeastModeMaths
@@BeastModeMaths Ah yes now I see, thanks.
Hey there, how do you do the sequence?
0,1,4,9.......
Thank you.
Hi BULL DOG,
I could go through the process that I went through in the video, but there is something that stands out...
The numbers you have listed are the square numbers. The 1st term is 0 squared, the 2nd term is 1 squared, the 3rd term is 2 squared and the 4th term is 3 squared... Each term is 1 less than the term number and then squared.
Therefore, the nth term would be (n-1)^2.
I hope that makes sense.
@@BeastModeMaths Thank you bunches. It totally made sense.
@@BeastModeMaths would it still be correct if my answer is n²-2+1?
@@emigo45 Yes that would be correct too. Your answer and my answer are identities (your one is just the expanded version of mine) 🙂
@@BeastModeMaths thank you very much
really good video
😄😃
Thanks!
really helped, easy step by step method on how to do it, thanks
You're very welcome. Thanks.
THANK YOU!!!!
You're welcome, Anna! 🙂
Hello, how do you do this sequence?
2,6,22,56,114,202
Thank you!
Hello Jane,
This is a cubic sequence, because the 3rd difference between terms is the same. Let me explain...
The 1st difference between terms is 4, 16, 34, 58, 88, ...
Therefore, the 2nd difference between terms is 12, 18, 24, 30, ...
Therefore, the 3rd difference between terms is 6, 6, 6, ...
Because we now know that it is a cubic sequence, it starts with n^3. To calculate the coefficient on the n^3 term, we divide the 3rd difference by 6. So, 6/6 is 1, so our nth term starts with 1n^3.
Now we can subtract the 1n^3 sequence away from our original sequence:-
Our sequence: 2, 6, 22, 56, 114, 202
1n^3 sequence: 1, 8, 27, 64, 125, 216
Our results is the following sequence: 1, -2, -5, -8, -11, -14.
Now we can calculate the nth term of this sequence and add it on to the cubic part that we just figured out. The nth term of the sequence above is -3n+4.
Combining everything together, we get n^3 - 3n + 4.
I hope that helps.
@@BeastModeMaths hii, how do you get +4?
So after we have taken the n^3 sequence away from our original sequence, we are left with the following sequence...
1, -2, -5, -8, -11, -14.
Now, if we calculate the nth term of this sequence, we know it must start with -3n (because it is going down by 3 each time, i.e. it is similar to our -3 x table). Now we just need to work out the adjustment. Can you see that each number is 4 places above our -3 times table? For example...
To go from -3 (1 x -3) to 1, we need to add 4
To go from -6 (2 x -3) to -2, we need to add 4
To go from -9 (3 x -3) to -5, we need to add 4
To go from -12 (4 x -3) to -8, we need to add 4
To go from -15 (5 x -3) to -11, we need to add 4
To go from -18 (6 x -3) to -14, we need to add 4
Just to summarise, the sequence above is the same as our -3 times tables, but each number has been shifted up by 4 places, hence -3n+4.
I hope that clears it up.
Thanks
For the knowledge
You are very welcome 🙂
How to take out term to term rule of quadratic sequence?
Hi Hasan Ali. For quadratic sequences, the term-to-term rule is different for each term. That is why we need to calculate the 2nd difference. If the second difference was +2, we could say that the term-to-term rule of this sequence was "add 2 more each time".
Thank you very much For helping that Quick
I had An assignment which I Was not able to understand but you made me understand with ease
Thanks 😊
thank you so so so much this was extremely helpful
Hi Sarah. You are very welcome 😊
what do you means by shifting up by 4, still dont get it
@@Jeysiiiiii I'm assuming you are referring to the example at the beginning , and the sequence of 6, 8, 10, 12, 14.
Notice that the numbers go up by 2 each time. Therefore, we can say that the sequence is similar to the 2 x table. If we compare it to the 2 x table, each number in our sequence is 4 units higher (e.g. 6 is 4 more than 2, 8 is 4 more than 4, 10 is 4 more than 6, etc.). So each number has been "shifted up" by 4 units. That's where we get the 2n+4 from (4 more than the 2 x table). I hope that helps!!
explained it so well, thank u
You're very welcome 🙂
Sir in 3,13,27,45,67 when your doing s term why did you put -3
Hello. The "-3" is part of the 4n-3 which is the nth term of the sequence 1, 5, 9, 13, 17.
Let me explain this. The numbers go up by 4 each time, which is where we get 4n from. If the sequence was just 4n then the numbers would go 4, 8, 12, 16, 20 (just the 4 x table). Notice that each number in our sequence is 3 less than the 4n sequence.
I hope that makes sense!
بارك الله فيك شرح رائع
Thanks
In the first q how did the n+3 came in ...? Can u explain it plzx
Hello Aaron,
After we subtracted the quadratic part of our sequence away from our original sequence, we were left with the sequence of numbers 4, 5, 6, 7, 8, ...
We then need to calculate what the nth term of this sequence is.
This sequence is similar to our 1 x table, but each number is just shifted up by 3 places
(for example, 4 is 1x1, plus 3)
(5 is 1x2, plus 3)
(6 is 1x3, plus 3)
(7 is 1x4, plus 3)
(8 is 1x5, plus 3)
So the nth term of this sequence of numbers is 1n+3.
Now, our final nth term is the quadratic part and the linear part put together...
Quadratic part = 2n^2
Linear part = n+3
Final nth term = 2n^2 + n +3
I hope that helps - I would recommend going back over calculating the nth term of linear sequences first if this part confused you.
@@BeastModeMaths thanku soo much sir it helped me so much😄😄
@@aaronnepali6459 You're welcome!
Thank you 🌹
You're welcome! I'm glad it was useful.
Awesome ❤
Tah
Hi how about this sequence 10, 13, 18, 25, 34?
Thank you!
Hello ba na na,
So, we first need to identify the 1st difference between terms - this is +3, +5, +7, +9.
Now we look at the 2nd difference - this is +2, +2, +2.
Because the 2nd difference is the same, we know that it is a quadratic sequence, so our nth term starts with __n^2. We just need to work out the number in front of the n^2 term. To do this, we just half the second difference. So half of 2 is 1. Therefore, our nth term starts with 1n^2 (or we could just say n^2).
Now we have found out the quadratic part of our sequence, we need to work out the linear part. Do do this, we just subtract our quadratic sequence (that we have just worked out) from our original sequence...
Original sequence: 10, 13, 18, 25, 34
n^2 sequence: 1, 4, 9, 16, 25
Difference: 9, 9, 9, 9, 9
Because the difference between the n^2 sequence and our original sequence is 9, we just need to add that on.
So our final nth term is n^2 + 9.
I hope all that helps!
Awesome.. thanks
You're very welcome!
last time you said that u need to half it but u didnt half it on 2:44
Hi Abishan Zi.
When we are finding the quadratic part of our sequence, we always half the second difference (for example, in the example I go through, the 2nd difference is 2, therefore half of 2 is 1, so it begins with 1n^2.
This is the only part where you half the 2nd difference. The part you are referring to at 2:44 is when calculating the nth term of an arithmetic sequence. If the numbers are 6, 8, 10, 12, 14, then the nth term of this sequence is 2n+4. If you are not quite sure about this part, I would recommend watching a video on how to calculate the nth term of an arithmetic sequence.
Many thanks.
Do u have to do the shifting?
Hi sfl,
The reason why we do the "shifting" is to make the sequence we are left with look the same as one of our times tables. This might not make much sense, but let me use the first example to help.
After we have found the quadratic part of our sequence and we have subtracted it from our original sequence, we are left with the numbers 6, 8, 10, 12, 14. This sequence is the same as our 2 times tables, but it has just been "shifted up" by 4. The only time that we would not need to "shift" would be if the sequence we are left with is exactly the same as one of our times tables (e.g. 2, 4, 6, 8, 10).
I hope that helps!
@@BeastModeMaths yes but I'm in year 9band have a test and im doing higher would i have to shift up or can I just leave it like dat
@@glamour1123 I don't know what questions will come up in your exam.
كيف يمكنني ايجاد formula لمتتالية تتكون من ارقام صحيحة موجبة وسالبة ؟
وايظا formula اخرى لمتتالية تتكون من ارقام كسرية موجبة وسالبة ؟
ارجو الاجابة لدي امتحان نهائي غدا 😢
How about this nth term sequence sir 2,8,18,32 ?
Hi Leonie.
So for this sequence
the first difference goes +6, +10, +14
then the second difference is +4, +4.
As the second difference is the same (+4), we half this number, so the quadratic part of our sequence starts with 2n^2.
Now we need to subtract the 2n^2 sequence away from our original sequence to see what is left over.
original sequence: 2, 8, 18, 32
2n^2 sequence: 2, 8, 18, 32
It turns out that the 2n^2 sequence is our sequence! We don't have to make any adjustment - so our nth term is just 2n^2
I hope that helps.
Thank you so much it really helped
_position_of_the___term _________/1,2,3,4,5,66
Term. /2,8,18,32 and under 66 the is nothing how do we work it out
@@leonietanke8048 I'm not sure exactly what you are asking. Are you trying to work out what the 66nd term of the sequence is?
If so, we can substitute 66 into our nth term...
= 2 x 66^2
= 2 x 4356
= 8712.
They gave me a table that show something like that so I was trying to make a table
Thank u so much
No problem! Glad to be able to help 🙂
Thanks
You're welcome!
But 5n+ 2 where is the +2 coming from bro
Hi yungogthathoop. Sorry for the late reply. I've only just seen your comment.
For the sequence 7, 12, 17, 22, 27, ... if we compare it with the 5 x table, notice how each number in our sequence is 2 bigger than the 5 x table:-
7 = 5 + 2
12 = 10 + 2
17 = 15 + 2
22 = 20 + 2 ...etc
That is where the +2 comes from.
I hope that helps!
How about the nth term of these sequence Sir , 0,10,24,56,112, 190?
Hi there. The list of numbers that you wrote down is not that of a typical sequence (it is neither arithmetic, geometric, quadratic nor Fibonacci).
@@BeastModeMaths by any chance do you have any idea what sequence are those? I've been confused how to answer it
@@jelmayguardiario8166 I have just calculated the "difference between the differences" and when you get to the fourth difference, it is the same. The 4th difference is -8, which means that this sequence could be a quartic sequence. I would not know how to calculate the nth term of this type of sequence though. Sorry I can't be of any more help.
@@BeastModeMaths still thank youuu Sir
1,5,25,125,3125 how do we solve it
Hi Leonie,
This is a geometric sequence because to go from one term to the next we multiply by the same number. For this sequence, to go from one term to the next, we multiply by 5.
We start at 1 and then x5 to get the next term (5)
We then x5 again to get the next term (25)
We then x5 again to get the next term (125)
*I think there is a mistake in the sequence that you gave - the next term should say 625 and then 3125 (I think you have just missed 625)
So to calculate our nth term, it will be 1 x 5^(n-1).
The power of 5 is always 1 less than our term number. The reason for this is because to get our 2nd term, we multiply by 5 once, to get our 3rd term, we multiply by 5 twice, and so on.
I have a video dedicated to geometric sequence, which should help!
Thank you its just that they are little bit confusing
Sir can you tell this please
Hello. Can I tell what? I'm confused 😕
Super
Thank you :)
What the fuck does shifting up by 3 mean
In the example you are referring to, the sequence that we are left with (after we have subtracted the quadratic sequence away from our original sequene) is 4, 5, 6, 7, 8. This sequence is the same as our 1 x table, but we have just added 3 to each number in the 1 x table. This is what I mean by being "shifted up by 3 places".
I hope that makes sense.
@@BeastModeMaths Oh, right it all makes sense now cheers
@@hystericalstratagem1434 You're welcome
Sir it took to long
Hi T9 HD. I would say that it doesn't matter too much if it takes a long time to do. The most important thing is doing it correctly so that you end up with the correct answer. Once you get more practice, it will soon become easier. Thanks for watching.