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Jonask
 one year ago
Best ResponseYou've already chosen the best response.0\[a_1=1/2\] and for each \[n \ge 2\] \[\huge a_n=\left(\frac{ 2n3 }{ 2n }\right)a_{n1}\] Prove that \[\huge\sum_{k=1}^{n}a_k<1 \] for all \[n \ge1\]

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1dw:1357738064697:dwJust try to Prove

Azteck
 one year ago
Best ResponseYou've already chosen the best response.1\[a_{1}+ a_{2}+…+ a_{n} < 1\] \[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2n  3 }{ 2n })a_{n1} < 1\] For n=1 LHS=\[\frac{ 1 }{ 2 }\] \[<RHS\] (Therefore) True for n=1 For n=2 LHS=\[(\frac{ 2(2)  3 }{ 2(2) })a_{21}\] \[<RHS\] (Therefore) True for n=2

Azteck
 one year ago
Best ResponseYou've already chosen the best response.1Assume the formula is true for n=m \[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2m  3 }{ 2m })a_{m1} < 1\] For n=m+1 \[\frac{ 1 }{ 2 } + a_{2} + a_{3}+…+(\frac{ 2m  3 }{ 2m })a_{m1}+(\frac{ 2(m+1)  3 }{ 2(m+1) })a_{m+11} < 1\] LHS= \[1 + (\frac{ 2(m+1)  3 }{ 2(m+1) })a_{m}\] \[=(\frac{ 2m1 }{ 2m+2 })(\frac{ 2m  3 }{ 2m })a_{m1}\]

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0@sauravshakya @azteck thanks

Jonask
 one year ago
Best ResponseYou've already chosen the best response.0not really worth 1 hour

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1I am not sure if this helps

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359039410279:dw

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1Did u mean like this @experimentX

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1yeah .. somewhat similar to that. but one terms is missing.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1there should be one n ... or i made mistake.

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1dw:1359040853679:dw

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.1It was me who did a mistake

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1hmm ... now lets use this identity. http://planetmath.org/GeneratingFunctionForTheCatalanNumbers.html
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