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May 5th, 2017, 10:18 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, 10:43 AM  #2 
Senior Member Joined: Feb 2010 Posts: 634 Thanks: 104  
May 5th, 2017, 10:48 AM  #3  
Math Team Joined: Jul 2011 From: Texas Posts: 2,699 Thanks: 1357  Quote:
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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