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If `(1 + x)^(n) = C_(0) + C_(1) x C_(2) x^(2) +…+ C_(n) x^(n)`, then the sum `C_(0) + (C_(0)+C_(1))+…+(C_(0) +C_(1) +…+C_(n -1))` is equal toA. `n 2^(n -1)`B. `( n-1)2^(n)`C. ` n 2 ^(n)`D. `(n+1)2^(n)` |
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Answer» Correct Answer - a We have, `C_(0) + (C_(0)+C_(1))+…+(C_(0) +C_(1) +C_(2)+…+C_(0)+C_(1) +C_(2)+...+ C_(n-1))` `nC_(0) +(n-1)C_(1) + (n-2) C_(2) +...+C_(n-1)` `= sum_(r=1)^(n)(n-r)""^(n)C_(r)` `= sum_(r=1)^(n)n. ""^(n)C_(r)- sum_(r=0)^(n) r. ""^(n)C_(r)` `=n sum_(r=1)^(n) ""^(n)C_(r)-n sum_(r=0)^(n) ""^(n-1)C_(r-1)= n2^(n) - n2^(n-1) = n2^(n-1)` . |
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