1 Answer

Iliiaz Akhmedov · Answer 1 · 2019-01-19T05:41:36+0000

2. If you already know Faulhaber's formula, which says

$\displaystyle\sum_(k=1)^nk^2=\frac{n(n+1)(2n+1)}6$

then it's just a matter of setting
n=4

. If you don't, then you can prove that it works (via induction), or compute the sum by some other means. Presumably you're not expected to use brute force and just add the squares of 1 through 4.

Just to demonstrate one possible method of verifying the formula, suppose we start from the binomial expansion of
(k-1)^3

, do some manipulation, then sum over
$1\le k\le n$ :

(k-1)^3=k^3-3k^2+3k-1

$\implies k^3-(k-1)^3=3k^2-3k+1$

$\implies\displaystyle\sum_(k=1)^n(k^3-(k-1)^3)=\sum_(k=1)^n(3k^2-3k+1)$

The left side is a telescoping series - several terms in consecutive terms of the series will cancel - and reduces to
n^3

. For example,

$\displaystyle\sum_(k=3)^2(k^3-(k-1)^3)=(1^3-0^3)+(2^3-1^3)+(3^3-2^3)=3^3$

Distributing the sum on the right side across each term and pull out constant factors to get

$\displaystyle n^3=3\sum_(k=1)^nk^2-3\sum_(k=1)^nk+\sum_(k=1)^n1$

If you don't know the formula for
$\displaystyle\sum_(k=1)^nk=\frac{n(n+1)}2$ , you can use a similar trick with the binomial expansion
(k-1)^2

, or a simpler trick due to Gauss, or other methods. I'll assume you know it to save space for the other parts of your question. We then have

$\displaystyle n^3=3\sum_(k=1)^nk^2-\frac{3n(n+1)}2+n$

$\implies\displaystyle\sum_(k=1)^nk^2=\frac{n(n+1)(2n+1)}6$

and when
n=4