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If the value of `"^(n)C_(0)+2*^(n)C_(1)+3*^(n)C_(2)+...+(n+1)*^(n)C_(n)=576`, then `n` isA. `7`B. `5`C. `6`D. `9` |
Answer» Correct Answer - A `(a)` `S=^(n)C_(0)+2.^(n)C_(1)+3.(n)C_(2)+...+(n+1).^(n)C_(n)` Here `T_(r )=(r+1)^(n)C_(r )=n^(n-1)C_(r-1)+^(n)C_(r )` `:.S=n^(2n-1)+2^(n)=576` (given) `:.n=7` |
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