In class we showed the following: n ∑(k=1) k = n(n+1)/2 and n∑(k=1) k^2 = n(n+ 1)(2n+ 1)/6 Using the fact that (k+ 1)^4-k^4= 4k^3+ 6k^2+ 4k+ 1 and summing up as k= 1,2,3,……, n together with the above two equalities, deduce that n∑(k=1) k^3 = (n(n+1)/2)^2

Let

$S_1 =\sum\limits_{k=1}^n1 = 1+1+1+...+1 = 1\cdot n = n$$S_2 =\sum\limits_{k=1}^nk=\frac{n(n+1)}{2}$$S_3 = \sum\limits_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$$S_4 = \sum\limits_{k=1}^nk^3$

Use the fact

$(k+1)^4 -k^4 =4k^3 +6k^2 +4k+1$

Put $k=1,2,3..., n$ to this formula:

$2^4 -1^4 =4\cdot 1^3+6\cdot 1^2 +4\cdot 1+1$

$3^4 -2^4 =4\cdot 2^3+6\cdot 2^2 +4\cdot 2+1$

$4^4 -3^4 =4\cdot 3^3+6\cdot 3^2 +4\cdot 3+1$

...

$(n+1)^4 -n^4 =4\cdot n^3+6\cdot n^2 +4\cdot n+1$

Sum left sides of these equalities:

$\cancel{2^4} -1^4 +\cancel{3^4} -\cancel{2^4} +\cancel{4^4} -\cancel{3^4} +...+(n+1)^4 -\cancel{n^4}=(n+1)^4-1$

Sum right sides of these equalities:

$4(1^3+2^3+...+n^3)+6(1^2+2^2+...+n^2)+4(1+2+...+n)+(1+1+...+1)=$

$=4S_4+6S_3 +4S_2 +S_1$

Hence

$4S_4+6S_3 +4S_2 +S_1 =(n+1)^4-1$

$4S_4=(n+1)^4-1-6S_3-4S_2-S_1$

$4S_4 =(n+1)^4 -1-\frac{6n(n+1)(2n+1)}{6}-\frac{4n(n+1)}{2}-n$

$4S_4 =(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n$

$4S_4 =(n+1)^4-n(n+1)(2n+1)-2n(n+1)-(n+1)$

$4S_4 =(n+1)((n+1)^3 -n(2n+1)-2n-1)$

$4S_4 =(n+1)(n^3+3n^2+\cancel{3n}+\cancel{1}-2n^2-\cancel{n}-\cancel{2n}-\cancel{1})$

$4S_4 =(n+1)(n^3+n^2)$

$4S_4 =(n+1)\cdot n^2(n+1)$

$4S_4 =n^2(n+1)^2$

$S_4 =\frac{n^2(n+1)^2}{4}=(\frac{n(n+1)}{2})^2$

$\sum\limits_{k=1}^nk^3 =(\frac{n(n+1)}{2})^2$