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Prove that `(.^(n)C_(1))/(2) + (.^(n)C_(3))/(4) + (.^(n)C_(5))/(6) + "…." = (2^(n) - 1)/(n+1)`. |
Answer» `S = (.^(n)C_(1))/(2)+(.^(n)C_(3))/(4)+(.^(n)C_(5))/(6)+"..."` `:. T_(r) = (.^(n)C_(2r-1))/(2r)= (.^(n+1)C_(2r))/(n+1)`, where `r = 1,2,3,"...."` `rArr S = 1/(n+1)(.^(n+1)C_(2)+.^(n+1)C_(4)+.^(n+1)C_(6)+"....")` ` = (1)/(n+1)[(.^(n+1)C_(0) + .^(n+1)C_(2) + .^(n+1)C_(4) +"....")-.^(n+1)C_(0)] = (2^(n)-1)/(n+1)`. |
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