J Pi you are the guy!!! Just finished my 3rd Oxford maths interview like 2 hours ago and all your videos on interview problems helped me so so much, especially with explaining my reasoning. No matter the result I'm super grateful for your help thank you!!!
@@ryankelly9772 amazing stuff! I'm sure you smashed it!!! You're welcome, and hopefully it all goes well!! Which colleges did you have interviews with?
Another nice way to solve it: imagine stacking up small cubes of side length one in a big cube of side length n. As n tends to infinity, the sum of squares tends to a pyramid, which has volume a third of the cube
As the integral from 0 to 1 of x^2 is smaller or equal to 1/n(xn), when you take the limit of n going to infinity, the rectangle's bases shrink down to zero, and the sum of their areas fit perfectly under the curve. That means that when you take the limit as n approaches infinity, 1/n(xn) fits on the Rieman sum definition of x^2 from 0 to 1. Therefore, the sum shall be exaclty 1^3/3 - 0^3/3 = 1/3
@@JPiMaths There's a very neat little trick that makes deriving these sum formulas relatively painless. The sums of the "incrementing" form k*(k+1)*(k+2)*...*(k+m-1) behave almost exactly like integrals of x^m. For example (The following sums are all from 1 to N): Sum(k) => N(N+1) / 2 Sum(k(k+1)) => N(N+1)(N+2) / 3 Sum(k(k+1)(k+2)) => N(N+1)(N+2)(N+3) / 4 ... It's almost the same as when you have an integral of powers of x^m: just add another x and divide by the exponent (the only difference is that you add (N+something) instead of just N, so that the form still remains "incrementing"). Here's how it would work for the sum of squares: We have Sum(k^2), which we need to write as a combination of the "incrementing" sums I mentioned. So, we can write it as: Sum(k^2) = Sum(k(k+1)) - Sum(k). Both of these are incrementing sums, and we can evaluate each one directly through the "integral like" rules. The first gives us N(N+1)(N+2) / 3, and the second gives us N(N+1) / 2. Then we just calculate the difference and get N(N+1)(2N+1) / 6. There's a very closely related tabular method (I think it's attributed to Newton and someone else, but can't remember who) which can easily shred through any such polynomial sum without much thinking.
J Pi you are the guy!!! Just finished my 3rd Oxford maths interview like 2 hours ago and all your videos on interview problems helped me so so much, especially with explaining my reasoning. No matter the result I'm super grateful for your help thank you!!!
@@ryankelly9772 amazing stuff! I'm sure you smashed it!!! You're welcome, and hopefully it all goes well!! Which colleges did you have interviews with?
@@JPiMaths Balliol and Catz 😁
@@ryankelly9772 ah nice! Let me know how it goes 🤞🤞🤞
Another nice way to solve it: imagine stacking up small cubes of side length one in a big cube of side length n. As n tends to infinity, the sum of squares tends to a pyramid, which has volume a third of the cube
Ah nice solve, I like this method!
As the integral from 0 to 1 of x^2 is smaller or equal to 1/n(xn), when you take the limit of n going to infinity, the rectangle's bases shrink down to zero, and the sum of their areas fit perfectly under the curve. That means that when you take the limit as n approaches infinity, 1/n(xn) fits on the Rieman sum definition of x^2 from 0 to 1. Therefore, the sum shall be exaclty 1^3/3 - 0^3/3 = 1/3
great videos!! i had my cambridge maths interviews last week and these videos were helpful in helping me think to use more abstract techniques 👍
@@jakehawkins2158 I'm glad! Hoping your interviews went smoothly! Which college was it at??
@ thank you!! it was at king’s college 👍
xD i just used lhopital
@vixguy I guess that works, but you still need to know how to simplify the sum, right?
@JPiMaths yes! :)
I did it the first way but had to derive the formula for the sum of the first n squares by hand (knowing it has to be some polynomial of degree three)
Ah yes, I can never remember the formula beyond the sum of cubes, but it's good to know how to derive it!
@@JPiMaths There's a very neat little trick that makes deriving these sum formulas relatively painless. The sums of the "incrementing" form k*(k+1)*(k+2)*...*(k+m-1) behave almost exactly like integrals of x^m. For example (The following sums are all from 1 to N):
Sum(k) => N(N+1) / 2
Sum(k(k+1)) => N(N+1)(N+2) / 3
Sum(k(k+1)(k+2)) => N(N+1)(N+2)(N+3) / 4
...
It's almost the same as when you have an integral of powers of x^m: just add another x and divide by the exponent (the only difference is that you add (N+something) instead of just N, so that the form still remains "incrementing"). Here's how it would work for the sum of squares:
We have Sum(k^2), which we need to write as a combination of the "incrementing" sums I mentioned. So, we can write it as:
Sum(k^2) = Sum(k(k+1)) - Sum(k). Both of these are incrementing sums, and we can evaluate each one directly through the "integral like" rules. The first gives us N(N+1)(N+2) / 3, and the second gives us N(N+1) / 2. Then we just calculate the difference and get N(N+1)(2N+1) / 6.
There's a very closely related tabular method (I think it's attributed to Newton and someone else, but can't remember who) which can easily shred through any such polynomial sum without much thinking.
Please keep posting them even after the admissions cycle ends! They are really interesting problems to just solve for fun!!!❤
@TM-ln5mr don't worry, I've got plenty more videos coming!
de tête :1/3.
trop facile !