My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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.
ahmed786 is offline  
 
May 5th, 2017, 09:43 AM   #2
Senior Member
 
mrtwhs's Avatar
 
Joined: Feb 2010

Posts: 621
Thanks: 98

Quote:
Originally Posted by ahmed786 View Post
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.
mrtwhs is offline  
May 5th, 2017, 09:48 AM   #3
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,579
Thanks: 1275

Quote:
Originally Posted by ahmed786 View Post
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 ...
skeeter is online now  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
induction, mathematical



Thread Tools
Display Modes


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





Copyright © 2017 My Math Forum. All rights reserved.