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Jonask
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1357738064697:dwJust try to Prove

Azteck
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0@sauravshakya @azteck thanks

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

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

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

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

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

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

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

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