The last one is great! since the first column's sum is n(n+1)/2 and the other columns are factors of the first one, then the total sum is n(n+1)/2 x (1+2+3 ...) = n(n+1)/2 x n(n+1)/2 = 45² = 2025. Hell yeah!
@@BanglaBhai49 sure… I am a B.Tech graduate in Electronics and Telecommunication engineering, 2008 batch. I’m an IT professional working as Microsoft Solutions Architect.
1:21 We all know that odd numbers from 1-89 = all numbers from 1-89 minus all even numbers from 1-89 so lets do that! All the numbers from 1-89 is 89(89+1)/2 which is equal to 4005 All even numbers from 1 to 89 is a little harder. But heres the solution. Just factor out the 2 from all of the numbers in the series and you get the sum of all numbers from 1 to n/2 multiplyed by 2. Thats the same thing as n/2 times n/2+1 devided by 2 and multiplyed by 2, and those cancel out. n in this situation would be 89, but 89 is odd and isnt in the series so we use 88. 88/2*(88/2+1) = 1980 4005-1980=2025 Youre welcome and a happy new year
Harshad number is a new thing you add it me. As far as sum of odd numbers upto 89 is easy . Take sum of 1 to 90 i.e 90*91/2 subtract even summation ( upto 88) from this i.e 2*45*46/2.
To validate the sum of the first 45 odd numbers. The sum of the first 45 odd numbers plus the sum of the first 45 even numbers is equivalent to the sum of the first 90 numbers. Hence the sum of the first 45 odd numbers is the sum of the 1st 90 numbers minus the first 45 even numbers. Let x = 45 hence 2x = 90. From the summing formula the sum of the 1st 90 numbers is 2x(2x+1)/2 which is 2x^2+x The sum of the first 45 even numbers is double the sum of the 45 numbers. This x(x+1)/2*2 which is x^2+x Subtract the final expressions from the above 2 lines and we are left with x^2 which is45^2,
Also, 2025 is among the first few elements in this self-referencing iterated sequence, via Domotro from Combo Class. Where T(n) is a triangular number, and with n > 1: T(2) = 3 3^2 = 9 T(9) = 45 45^2 = 2,025 T(2,025) = 2,051,325 2,051,325^2 = 4,207,934,255,625 T(4,207,934,255,625) = 8,853,355,349,833,265,389,198,125 Tn^2 = 78,381,900,950,421,300,982,881,904,787,876,752,731,430,503,515,625 k = sum of first n cubes n = 2, k = 9 n = 9, k = 2,025 n = 2,025, k = 4,207,934,255,625 n = above, k = 78,381,900,950,421,300,982,881,904,787,876,752,731,430,503,515,625
2025 is the square of 45 which happens to be the sum of 1 to 9, much of the rest are properties that are hardly unique to 2025. All perfect squares are the sum of a series of odd numbers, all numbers whose digits add up to 9 are divisible by 9, and the sum of the cubes from 1 to whatever number n is always the sum of 1 to n squared.
The last one is great! since the first column's sum is n(n+1)/2 and the other columns are factors of the first one, then the total sum is n(n+1)/2 x (1+2+3 ...) = n(n+1)/2 x n(n+1)/2 = 45² = 2025. Hell yeah!
edit: you can already shoot and pre-publish the 3025 happy new year video, this will be the sum of usual times tables that go up to 10.
@@cheeseparis1 TH-cam might not be existing at that time😑
@@cheeseparis1 wooohhhh.. that was good
If you don't mind sir may I know your educational background?@@LOGICALLYYOURS
@@BanglaBhai49 sure… I am a B.Tech graduate in Electronics and Telecommunication engineering, 2008 batch. I’m an IT professional working as Microsoft Solutions Architect.
1:21
We all know that odd numbers from 1-89 = all numbers from 1-89 minus all even numbers from 1-89 so lets do that!
All the numbers from 1-89 is 89(89+1)/2 which is equal to 4005
All even numbers from 1 to 89 is a little harder. But heres the solution. Just factor out the 2 from all of the numbers in the series and you get the sum of all numbers from 1 to n/2 multiplyed by 2. Thats the same thing as n/2 times n/2+1 devided by 2 and multiplyed by 2, and those cancel out. n in this situation would be 89, but 89 is odd and isnt in the series so we use 88. 88/2*(88/2+1) = 1980
4005-1980=2025
Youre welcome and a happy new year
This is one of the two solutions I had in my mind…. I really appreciate your comment ✌️
sum of first n odd numbers is n^2
Perfect
Really cool stuff! Well done on collating all the patterns into one video.
As a heads up, 7. is a direct consequence of 4. if you expand the brackets.
Thanks for the appreciation… and a great observation 😊
Harshad number is a new thing you add it me. As far as sum of odd numbers upto 89 is easy . Take sum of 1 to 90 i.e 90*91/2 subtract even summation ( upto 88) from this i.e 2*45*46/2.
Amazing patterns bro ❤❤❤
Thanks for the love! I'm really glad you liked it.
To validate the sum of the first 45 odd numbers.
The sum of the first 45 odd numbers plus the sum of the first 45 even numbers is equivalent to the sum of the first 90 numbers. Hence the sum of the first 45 odd numbers is the sum of the 1st 90 numbers minus the first 45 even numbers.
Let x = 45 hence 2x = 90.
From the summing formula the sum of the 1st 90 numbers is 2x(2x+1)/2 which is 2x^2+x
The sum of the first 45 even numbers is double the sum of the 45 numbers. This x(x+1)/2*2 which is x^2+x
Subtract the final expressions from the above 2 lines and we are left with x^2 which is45^2,
happy new year 45^2
😁 logical way to wish
Also, 2025 is among the first few elements in this self-referencing iterated sequence, via Domotro from Combo Class.
Where T(n) is a triangular number,
and with n > 1:
T(2) = 3
3^2 = 9
T(9) = 45
45^2 = 2,025
T(2,025) = 2,051,325
2,051,325^2 = 4,207,934,255,625
T(4,207,934,255,625) = 8,853,355,349,833,265,389,198,125
Tn^2 = 78,381,900,950,421,300,982,881,904,787,876,752,731,430,503,515,625
k = sum of first n cubes
n = 2, k = 9
n = 9, k = 2,025
n = 2,025, k = 4,207,934,255,625
n = above, k = 78,381,900,950,421,300,982,881,904,787,876,752,731,430,503,515,625
Wonderful coincidences for 2025❤
Interestingly, it seems all of these properties except the second one are shared by the last sum of cubes, 1296
That's a great observation. I'll have to check that out!
First Comment. Love your videos and puzzles.
Thanks Krishna 🤩
You missed two:
1. 9²x5²
2. 40² +20² + 5²
2025 is the square of 45 which happens to be the sum of 1 to 9, much of the rest are properties that are hardly unique to 2025. All perfect squares are the sum of a series of odd numbers, all numbers whose digits add up to 9 are divisible by 9, and the sum of the cubes from 1 to whatever number n is always the sum of 1 to n squared.
n^2 = 45^2
Perfect!… To those who didn’t get this, it is one of the ways to validate how we got the sum of first 45 odd numbers.
[(2+0+2+5)(2+0-2+5)]^2 = 2025
Solv blank sudoku in 4-5 mintues i can can u
Toughest puzzle
@@allroundvideosetc please share more details. Do you want me to make a video on it
20^2 400
25^2 625
+ 1025
(20+25)^2 and 20^2+25^2 aren't the same. (Binomial formula)