Tuesday, 10 June 2014

summation - Find the sum of the series of frac11cdot3+frac13cdot5+frac15cdot7+frac17cdot9+...



Find the sum of the series
113+135+157+179+1911+
My attempt solution:
13(1+15)+17(15+19)+111(19+113)+
=13(65)+17(1445)+111(22117)+
=2((15)+(145)+(1117)+)

=2((15)+(159)+(1913)+)
It is here that I am stuck. The answer should be 12 but I don't see how to get it. Any suggestions?



Also, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills?


Answer



This is a general approach to evaluate the sum of series, like these.



First find nth term of series.



Let Tn denote the nth term.




We see that,



T1=113



T2=135



And so on.
Let the numbers in green be
X1,X2,X3,X4,..=1,3,5,7...




Clearly they form an A.P. with common difference =2



So, nth term of this AP is 1+(n1)×2=2n1



Similarly,
Let the numbers in teal be Y1,Y2,Y3,Y4,..=3,5,7,9...
Clearly they form an A.P. with common difference =2



So, nth term of this AP is 3+(n1)×2=2n+1




So, the nth term of the main question is just



Tn=1(2n1)(2n+1)



Now, taking summation from 1 to n , we have,



nn=11(2n1)(2n+1)



=nn=1=12(2n+1)(2n1)(2n1)(2n+1)




=nn=1=121(2n1)1(2n+1)



=nn=1=12(11(2n+1))



While n=,



n=1=12(11(2()+1))



=n=1=12(11)




Since 1=0,



n=1=12(10)



Which is



n=1=12


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