August 18th, 2015, 10:15 AM  #1 
Member Joined: Aug 2015 From: Podgorica Posts: 61 Thanks: 0  Induction
Hello, can you help me solve or at least explain this problem? Yesterday I have started practicing Inductions,sequences and series but today it seems that I got stuck at this problem. Result: A)124 B)136 C)144 D)158 E)162 Last edited by med1student; August 18th, 2015 at 10:28 AM. 
August 18th, 2015, 12:40 PM  #2 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
$\displaystyle =\sum_{m=1}^3\sum_{k=0}^2((k+2+m)+(k+3+m)+(k+4+m)) =$, collect like terms, $\displaystyle =\sum_{m=1}^3\sum_{k=0}^2(3k+3m+9)=$ $\displaystyle =\sum_{m=1}^3((3m+9)+(3+3m+9)+(6+3m+9))=$, collect like terms, ... =162 
August 18th, 2015, 03:16 PM  #3 
Senior Member Joined: Aug 2014 From: United States Posts: 136 Thanks: 21 Math Focus: Learning 
Wait what induction?

August 18th, 2015, 05:13 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,422 Thanks: 1462 
$\displaystyle \sum_{i=2}^4\sum_{m=1}^3\sum_{k=0}^2k + \sum_{m=1}^3\sum_{k=0}^2\sum_{i=2}^4i + \sum_{k=0}^2\sum_{i=2}^4\sum_{m=1}^3m = 9(0 + 1 + 2) + 9(2 + 3 + 4) + 9(1 + 2 + 3) = 162$

August 19th, 2015, 06:59 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,921 Thanks: 785  Are you clear on what "induction" is? It is a technique for proving that a given statement is true for all positive integers. Since this problem allows m and k to take on only a finite number of values, induction is not appropriate. In fact, since that 'finite number of values' is very small (1, 2, and 3 for m, 0, 1, and 2 for k) there is no reason at all not to right out all the terms as skaa and skipjack did. 
August 19th, 2015, 08:11 AM  #6 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
I always though that induction is a method of mathematical proof, so... I think the problem is not related to induction. 

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