Induction

**rugbygirl2** · November 6th 2008, 09:47 AM

I need help with a mathematical induction proof to prove that
(1^2)+(3^2)+(5^2)+.....+(2n+1)^2 = ((n+1)(2n+1)(2n+3))/3 for all integers n is greater than or equal to zero.

I understand induction but I can't figure out the algebra, any help would be appreciated! Thanks.

**Soroban** · November 6th 2008, 11:23 AM

Hello, rugbygirl2!

Evidently, this sequence begins with $n = 0$

Prove: . $1^2+3^2+5^2+ \hdots + (2n+1)^2 \;=\;\frac{(n+1)(2n+1)(2n+3)}{3}$

Verify $S(1)\!:\;\;1^2+3^2 \:=\:\frac{2\cdot3\cdot5}{3}$ . . . True!

Assume $S(k)\!:\;\;1^2+3^2+5^2 + \hdots + (2k+1)^2 \;=\;\frac{(k+1)(2k+1)(2k+3)}{3}$

Add $(2k+3)^2$ to both sides:

. . $\underbrace{1^2 + 3^2 + 5^2 + \hdots + (2k+3)^2}_{\text{The left side of }S(k+1)} \;=\;\frac{(k+1)(2k+1)(2k+3)}{3} + (2k+3)^2$

We must show that the right side is the right side of $S(k+1)$
. . which is: . $\frac{\bigg[(k+1)+1\bigg]\bigg[2(k+1)+1\bigg]\bigg[2(k+1)+3\bigg]}{3} \;=\;\frac{(k+2)(2k+3)(2k+5)}{3}$ .[1]

The right side is: . $\frac{(k+1)(2k+1)(2k+3)}{3} + (2k+3)^2$

Factor: . $\frac{2k+3}{3}\,\bigg[(k+1)(2k+1) + 3(2k+3)\bigg] \;=\;\frac{2k+3}{3}\,(2k^2 + 3k + 1 + 6k + 9)<br />$

. . . . $= \;\frac{2k+3}{3}\,(2k^2 + 9k+10) \;=\;\frac{2k+3}{3}\,(k+2)(2k+5)$

. . . . $= \;\frac{(k+2)(2k+3)(2k+5)}{3}\quad\hdots$ . which equals [1]

Our inductive proof is complete . . .

**rugbygirl2** · November 6th 2008, 08:07 PM

Thank you so much for your help!!!!!!!!!!!!!

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