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shubhamsrg
Group Title
infinite summation :
1 + (1/3) + (1.3 / 3.6) + (1.3.5 / 3.6.9) + (1.3.5.7 / 3.6.9.12) ........ = ?
 one year ago
 one year ago
shubhamsrg Group Title
infinite summation : 1 + (1/3) + (1.3 / 3.6) + (1.3.5 / 3.6.9) + (1.3.5.7 / 3.6.9.12) ........ = ?
 one year ago
 one year ago

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shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
that's central binomial coefficient ... google for something like it.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
it's better to use the generating function directly http://mathworld.wolfram.com/images/equations/CentralBinomialCoefficient/NumberedEquation1.gif
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
didnt get you ?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
http://mathworld.wolfram.com/CentralBinomialCoefficient.html put x=1/6 and get it's value
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
ohh,,nice !! where does this come from ?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1349165062856:dw
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
but its C(2n,n) ?!?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
yeah ... I'm just playing around. I've seen that method too ... kinda forget it though.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
hmm,,so i should just rote it for the time being !! hmm..anyways ,,thank you.. :)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
should i just tag @mukushla @eliassaab
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
dw:1349165603145:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
probably it comes from what google said \[ (1+ax)^k = \sum_{i=0}^k\binom{k}{i}(ax)^i = \sum_{n=0}^\infty \binom{2n}{n}x^k\] since it's infinite k can't be natural number.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
have to prove that a=4 and k=1/2
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
here is one \[ 1 + {1 \over 3} + {1 \cdot 3 \over 3\cdot 6} + {1\cdot3\cdot 5\over 3\cdot6\cdot9}+...\] \[ (1 + x)^n = 1 + nx + {n(n1) \over 1 \cdot2} x^2 + ....+\infty \] assuming n to be non natural \[ nx = {1 \over 3}\\ {n(n1)\over 1 \cdot2}x^2 = {1 \over 6} \\ \text{ this gives } x={2 \over 3} , n={1 \over 2}\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
\[ \left(1  {2 \over 3}\right)^{1 \over 2} = \sqrt{3}\]
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
hmm..i get the idea i guess.//thanks !! :)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.3
there is bigger idea http://math.stackexchange.com/questions/205898/howtoshowthat1oversqrt14xgeneratesumn0inftybinom2n
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
ohh yes,,gotcha,,that was really helpful..thanks a ton sir!
 one year ago
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