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`Sigma_(r=1)^(n) (-1)^(r+1).(""^(n)C_(r))/(r+1) " is equal to" -`A. `-(1)/(n)`B. `(1)/(n+1)`C. `(-1)/(n+1)`D. `(n)/(n+1)` |
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Answer» Correct Answer - D `(1)/(n+1) overset(n)underset(r=1) Sigma (-1)^(r+1).(n+1)/(r+1).^(n)C_(r)` `=(1)/(n+1) overset(n)underset(r=1)Sigma(-1)^(r+1 " "n+1)C_(r+1)` `=(1)/(n+1)[.^(n+1)C_(2^(-)).^(n+1)C_(3)+.^(n+1)C_(4)-,,,+(-1)^(n+1" "n+1)C_(n+1)]` `=(1)/(n+1)[.^(n+1)C_(0)-^(n+1)C_(1)+^(n+1)C_(2)-,,,+(-1)^(n+1" "n+1)C_(n+1)+n]` `=(1)/(n+1)[0 +n] =(n)/(n+1)` |
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