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## Here's the Solution to this Question

Let $P(n)$ be the proposition that for

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

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

for all nonnegative integers $n.$

BASIS STEP:

$P(0)$is true because

$(2(0)+1)^2=1=\dfrac{(0+1)(2(0)+1)(2(0)+3)}{3}$

This completes the basis step.

INDUCTIVE STEP: For the inductive hypothesis, we assume that $P(k)$ is true for an arbitrary nonnegative integer $k.$ That is, we assume that

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

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

To carry out the inductive step using this assumption, 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

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

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

assuming the inductive hypothesis $P(k).$ Under the assumption of $P(k),$ we see that

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

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

$=\dfrac{(2k+3)(2k^2+3k+1+6k+9)}{3}$

$=\dfrac{(2k+3)(2k^2+9k+10)}{3}$

$=\dfrac{(2k+3)(2k+5)(k+2)}{3}$

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

Note that we used the inductive hypothesis in the second equation in this string of equalities to replace $1^2+3^2+5^2+...+(2k+1)^2$ by $\dfrac{(k+1)(2k+1)(2k+3)}{3}.$ We have completed the inductive step.

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

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

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

for all nonnegative integers $n.$