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anonymous
 3 years ago
\[\lim_{n \rightarrow \infty} \frac{ [\sum_{r=1}^{2n1 }(sr1 ]^3 }{[ \sum_{r=1}^{2n1}(sr1)^2]^2 }\]
anonymous
 3 years ago
\[\lim_{n \rightarrow \infty} \frac{ [\sum_{r=1}^{2n1 }(sr1 ]^3 }{[ \sum_{r=1}^{2n1}(sr1)^2]^2 }\]

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experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0what the hell is that??

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0exand the summation inside ,, its arithmetic series, and constant. express it in compact form and do business with cube and square outside.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0What is the difference between numerator and denom ? except for sq. and cube.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \frac{ [\sum_{r=1}^{2n1 }(sr1 ]^3 }{[ \sum_{r=1}^{2n1}(sr1)^2]^2 }\]

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0[ (s1) + (2s1) +(3s1) ...((2n1)s 1) ]^3  [ (s1)^2 + (2s1)^2 ...... ((2n1)s 1)^2 ]^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sr  = s1 , s2 , s3 , s4 ...................

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0numerator = [ s + 2s +3s .. (2n1)s  (2n1) ]^3 denominator = [ s^2 + (2s^2) ..((2n1)s)^2  2s(1+2+3..(2n1)) +(2n1) ]^2 This should help ?
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