1. ## Mathematical Induction

Prove, by mathematical induction, that

(1^3) + (2^3) + (3^3) + ... + (n^3) = ((n(n+1))/2)^2 for all integers n≥1
I got to here (below) and am stuck...
(1^3) + (2^3) + (3^3) + ... + ((n+1)^3) = (((n(n+1))/2)^2) + (n+1)^3

2. Originally Posted by bearej50
Prove, by mathematical induction, that

(1^3) + (2^3) + (3^3) + ... + (n^3) = ((n(n+1))/2)^2 for all integers n≥1
I got to here (below) and am stuck...
(1^3) + (2^3) + (3^3) + ... + ((n+1)^3) = (((n(n+1))/2)^2) + (n+1)^3
Hi bearej50,

P(k)

$\displaystyle 1^3+2^3+3^3+.....+k^3=\left[\frac{k(k+1)}{2}\right]^2$ ?

P(k+1)

$\displaystyle 1^3+2^3+3^3+.....+k^3+(k+1)^3=\left[\frac{(k+1)(k+2)}{2}\right]^2$

$\displaystyle 1^3+2^3+3^3+.....+k^3+(k+1)^3=\left[\frac{k(k+1)}{2}\right]^2+(k+1)^3$

if the first statement is true, and this must equal $\displaystyle \left[\frac{(k+1)(k+2)}{2}\right]^2$

Proof

$\displaystyle \left[\frac{k(k+1)}{2}\right]^2+(k+1)^3=(k+1)^2\frac{k^2}{4}+(k+1)^2(k+1)$

$\displaystyle =(k+1)^2\left(\frac{k^2}{4}+k+1\right)=(k+1)^2\lef t(\frac{k^2+4k+4}{4}\right)$

$\displaystyle =\frac{(k+1)^2(k+2)^2}{4}=\left[\frac{(k+1)(k+2)}{2}\right]^2$

hence P(k+1) is valid if P(k) is,

hence test for an initial value N.