infinite summation :

1 + (1/3) + (1.3 / 3.6) + (1.3.5 / 3.6.9) + (1.3.5.7 / 3.6.9.12) ........ = ?

Question

infinite summation :

1 + (1/3) + (1.3 / 3.6) + (1.3.5 / 3.6.9) + (1.3.5.7 / 3.6.9.12) ........ = ?

shubhamsrg · Answer

so far, my progress has been this much :
i started my series from (1/3) ... 
denom. of nth term is easily (3^n)*(n!)
num. of nth term will be
1.3.5.7............
=(1.2.3.4.5...........)/(2.4.6.8........)
=( (2n)! )/(2^n  * n! )

thus nth term will be 
(2n)! / (n!)^2 * 6^n 
or C(2n,n) / 6^n

we have to fine its summation from n = 1 to infinity..

but am clueless after this..
will this be even helpful ?

experimentx · Answer

that's central binomial coefficient ... google for something like it.

shubhamsrg · Answer

well i have googled over,,i got this solution,,but couldnt understand as to why we did that..
it said to compare the summation with (1+x)^n

and solve ultimately,,and to my surprise, i got the ans.. if anyone can explain why ?

experimentx · Answer

it's better to use the generating function directly http://mathworld.wolfram.com/images/equations/CentralBinomialCoefficient/NumberedEquation1.gif

shubhamsrg · Answer

didnt get you ?

experimentx · Answer

http://mathworld.wolfram.com/CentralBinomialCoefficient.html put x=1/6 and get it's value

shubhamsrg · Answer

ohh,,nice !! where does this come from ?

experimentx · Answer

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