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If `C_(0), C_(1), C_(2), ..., C_(n)` denote the binomial cefficients in the expansion of `(1 + x )^(n)` , then ` a C_(0) + (a + b) C_(1) + (a + 2b) C_(2) + ...+ (a + nb)C_(n) = `.A. `(a + nb ) 2^(n)`B. `(2a + nb) 2^(n)`C. `(a + nv)2^(n-1)`D. `(2a + nb) 2^(n-1)` |
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Answer» Correct Answer - d We have, ` a C_(0) + (a + b) C_(1) + (a + 2b) C_(2) + ...+ (a + nb)C_(n) ` `sum_(r=0)^(n) (a + rb)""^(n)C_(r)` `sum_(r=0)^(n) a.""^(n)C_(r) + sum_(r=0)^(n) rb""^(n)C_(r)` `=a(sum_(r=0)^(n) ""^(n)C_(r)) +b (sum_(r=0)^(n) r""^(n)C_(r))` `=a(sum_(r=0)^(n) ""^(n)C_(r)) +b (sum_(r=0)^(n) r.(n)/(r)""^(n-1)C_(r-1))` `=a(sum_(r=0)^(n) ""^(n)C_(r)) +b n(sum_(r=0)^(n) ""^(n-1)C_(r-1))` `a. 2^(n) + bn2^(n-1) " " [because sum_(r=0)^(n) ""^(n)C_(r)=2^(n), sum_(r=1)^(n) ""^(n-1)C_(r-1)= 2^(n-1)]` = `(2a + bn) 2^(n-1)` . |
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