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shubhamsrg

  • 2 years ago

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

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  1. shubhamsrg
    • 2 years ago
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    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 ?

  2. experimentX
    • 2 years ago
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    that's central binomial coefficient ... google for something like it.

  3. shubhamsrg
    • 2 years ago
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    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 ?

  4. experimentX
    • 2 years ago
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    it's better to use the generating function directly http://mathworld.wolfram.com/images/equations/CentralBinomialCoefficient/NumberedEquation1.gif

  5. shubhamsrg
    • 2 years ago
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    didnt get you ?

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

  7. shubhamsrg
    • 2 years ago
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    ohh,,nice !! where does this come from ?

  8. experimentX
    • 2 years ago
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    |dw:1349165062856:dw|

  9. shubhamsrg
    • 2 years ago
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    but its C(2n,n) ?!?

  10. experimentX
    • 2 years ago
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    yeah ... I'm just playing around. I've seen that method too ... kinda forget it though.

  11. shubhamsrg
    • 2 years ago
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    hmm,,so i should just rote it for the time being !! hmm..anyways ,,thank you.. :)

  12. shubhamsrg
    • 2 years ago
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    should i just tag @mukushla @eliassaab

  13. experimentX
    • 2 years ago
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    |dw:1349165603145:dw|

  14. experimentX
    • 2 years ago
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    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.

  15. experimentX
    • 2 years ago
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    have to prove that a=-4 and k=-1/2

  16. experimentX
    • 2 years ago
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    here is one \[ 1 + {1 \over 3} + {1 \cdot 3 \over 3\cdot 6} + {1\cdot3\cdot 5\over 3\cdot6\cdot9}+...\] \[ (1 + x)^n = 1 + nx + {n(n-1) \over 1 \cdot2} x^2 + ....+\infty \] assuming n to be non natural \[ nx = {1 \over 3}\\ {n(n-1)\over 1 \cdot2}x^2 = {1 \over 6} \\ \text{ this gives } x={-2 \over 3} , n={-1 \over 2}\]

  17. experimentX
    • 2 years ago
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    \[ \left(1 - {2 \over 3}\right)^{-1 \over 2} = \sqrt{3}\]

  18. shubhamsrg
    • 2 years ago
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    hmm..i get the idea i guess.//thanks !! :)

  19. experimentX
    • 2 years ago
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    there is bigger idea http://math.stackexchange.com/questions/205898/how-to-show-that-1-over-sqrt1-4x-generate-sum-n-0-infty-binom2n

  20. shubhamsrg
    • 2 years ago
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    ohh yes,,gotcha,,that was really helpful..thanks a ton sir!

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