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 May 5th, 2017, 09:18 AM #1 Newbie   Joined: May 2017 From: Pakistan Posts: 7 Thanks: 0 Mathematical Induction Prove the following by Mathematical Induction step by step 2²+4²+6²+......+(2n)²=3/2n(n+1)(2n+1), for all natural numbers n.
May 5th, 2017, 09:43 AM   #2
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Quote:
 Originally Posted by ahmed786 Prove the following by Mathematical Induction step by step 2²+4²+6²+......+(2n)²=3/2n(n+1)(2n+1), for all natural numbers n.
This formula is incorrect.

May 5th, 2017, 09:48 AM   #3
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Quote:
 Originally Posted by ahmed786 Prove the following by Mathematical Induction step by step 2²+4²+6²+......+(2n)²=3/2n(n+1)(2n+1), for all natural numbers n.
the correct sum is $\displaystyle \sum_{k=1}^n (2k)^2 = \dfrac{2}{3} \cdot n(n+1)(2n+1)$

true for $n=1$

$\displaystyle \sum_{k=1}^{n+1} (2k)^2 = \dfrac{2}{3} \cdot n(n+1)(2n+1) + [2(n+1)]^2$

you need to show $\dfrac{2}{3} \cdot n(n+1)(2n+1) + [2(n+1)]^2 = \dfrac{2}{3} \cdot (n+1)(n+2)(2n+3)$ ...

$\dfrac{2}{3} \cdot n(n+1)(2n+1) + [2(n+1)]^2 =$

$\dfrac{2}{3} \cdot n(n+1)(2n+1) + 4(n+1)^2 =$

$2(n+1) \bigg[\dfrac{n(2n+1)}{3} + 2(n+1)\bigg] =$

$2(n+1) \bigg[\dfrac{n(2n+1)}{3} + \dfrac{6(n+1)}{3}\bigg] =$

$\dfrac{2(n+1)}{3} \bigg[n(2n+1) + 6(n+1)\bigg] =$

$\dfrac{2(n+1)}{3} \bigg[2n^2+7n+6 \bigg] =$

finish it ...

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