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## Here's the Solution to this Question

Let $P(n)$ be the proposition that $1^3+2^3+...+n^3=(\dfrac{n(n+1)}{2})^2$ for the positive integer $n.$

(a) $P(1)$ is the statement that $1^3=(\dfrac{1(1+1)}{2})^2.$

(b) Basic Step

$P(1)$ is true because

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

This completes the basis step.

(c) For the inductive hypothesis, we assume that $P(k)$ is true for an arbitrary

positive integer $k.$ That is, we assume that

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

(d) Inductive Step

To carry out the inductive step we must show that when we assume that $P(k)$ is true, then $P(k + 1)$ is also true. That is, we must show that

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

assuming the inductive hypothesis $P(k).$

(e) Under the assumption of $P(k),$ we see that

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

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

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

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

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

Note that we used the inductive hypothesis in the second equation in this chain of equalities to replace $1^3+2^3+...+k^3$ with$(\dfrac{k(k+1)}{2})^2.$

We have completed the inductive step.

(f) Because we have completed the basis step and the inductive step, by mathematical induction we know that $P(n)$ is true for all positive integers $n.$ That is, $1^3+2^3+...+n^3=(\dfrac{n(n+1)}{2})^2$ for the positive integer $n.$

5. Let $P(n)$ be the proposition that

$1^2+3^2+5^2+...+(2n+1)^2$