Answer to Question #349929 in Discrete Mathematics for jean

Question

let p(n) 1^3+2^3+....+n^3=(n(n+1)/2)2 for positive integer n

Answer 1

Let $P(n)$ be the proposition that the sum of the first $n$ positive integers,

$$1^3+2^3+....+n^3=(\dfrac{n(n+1)}{2})^2$$

Basis Step

$P(1)$ is true because $1^3=1=(\dfrac{1(1+1)}{2})^2.$

Inductive Step

For the inductive hypothesis we assume that $P(k)$ holds for an arbitrary positive integer $k.$ That is, we assume that

$$1^3+2^3+....+k^3=(\dfrac{k(k+1)}{2})^2$$

Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that

$$1^3+2^3+....+k^3+(k+1)^3=(\dfrac{(k+1)(k+1+1)}{2})^2$$

is also true.

When we add $k + 1$ to both sides of the equation in $P(k),$ we obtain

$$1^3+2^3+....+k^3+(k+1)^3=$$

$$=(\dfrac{k(k+1)}{2})^2+(k+1)^3$$

$$=(\dfrac{k+1}{2})^2(k^2+4k+4)$$

$$=(\dfrac{k+1}{2})^2(k+2)^2$$

$$=(\dfrac{(k+1)(k+1+1)}{2})^2$$

This last equation shows that $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n.$ That is, we have proved that

$$1^3+2^3+....+n^3=(\dfrac{n(n+1)}{2})^2$$

for all positive integers $n.$