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Prove that (.^(n)C(0))/(1)+(.^(n)C(2))/(...

Prove that `(.^(n)C_(0))/(1)+(.^(n)C_(2))/(3)+(.^(n)C_(4))/(5)+(.^(n)C_(6))/(7)+"....."+= (2^(n))/(n+1)`

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`S = (.^(n)C_(0))/(1)+(.^(n)C_(2))/(3)+(.^(n)C_(4))/(5)+(.^(n)C_(6))/(7)+"….."`
The general term of the series in
`(.^(n)C_(2r))/(2r+1) = (.^(n+1)C_(2r+1))/(n+1)`, where `r = 0,1,2,"….."`
`:. S = (1)/(n+1)[.^(n+1)C_(1)+.^(n+1)C_(3)+.^(n+1)C_(5)+"......."] = (2^(n))/(n+1)`
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Prove that (.^(n)C_(1))/(2) + (.^(n)C_(3))/(4) + (.^(n)C_(5))/(6) + "…." = (2^(n) - 1)/(n+1) .

if (1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a++.^(n)C_(2)a^(2)+ . . .+.^(n)C_(n)a^(n) , then prove that (.^(n)C_(1))/(.^(n)C_(0))+(2(.^(n)C_(2)))/(.^(n)C_(1))+(3(.^(n)C_(3)))/(.^(n)C_(2))+. . . +(n(.^(n)C_(n)))/(.^(n)C_(n-1))= Sum of first n natural numbers.

(.^(n)C_(1))/(2)-(2(.^(n)C_(2)))/(3)+(3(.^(n)C_(3)))/(4)-....+(-1)^(n+1)(n(.^(n)C_(n)))/(n+1)=

(*^(n)c_(0))^(2)+3(.^(n)C_(1))^(2)+5(.^(n)C_(2))^(2)+ ......+(2n+1)(.^(n)C_(n))^(2)