My Math Forum Mathematical Induction
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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
Senior Member

Joined: Feb 2010

Posts: 610
Thanks: 93

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
Math Team

Joined: Jul 2011
From: Texas

Posts: 2,518
Thanks: 1240

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 ...

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Lambert1 Abstract Algebra 3 November 7th, 2015 06:33 AM creatingembla Calculus 3 October 19th, 2015 09:46 AM medos Algebra 5 October 31st, 2012 03:54 PM MathematicallyObtuse Algebra 2 February 5th, 2011 04:59 PM remeday86 Applied Math 1 June 20th, 2010 09:52 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top