Answer to Question #285297 in Discrete Mathematics for ali

Question

show that

1²+3²⁺5²+.....(2n+1)²=(n+1)(2n+1)(2n+3)/3

where n is a nonnegative integer.

Answer 1

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 $n.$