InterviewSolution
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InterviewSolution
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If `C_(0), C_(1), c_(2),...,C_(n)` denote the binomial coefficients in the expansion of `(1 + x)^(n)`, then `C_(0) + (C_(1))/(2) + C_(2)/(3) + ...+ (C_(n))/(n+1) or, sum_(r=0)^(n) (C_(r))/(r+ 1)`A. `(2^(n+1) + 1)/(n+ 1)`B. `(2^(n+1) - 1)/(n+ 1)`C. `(2^(n) + 1)/(n+ 1)`D. `(2^(n) - 1)/(n+ 1)` |
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Answer» Correct Answer - b `C_(0) + (C_(1))/(2) + C_(2)/(3) + ...+ (C_(n))/(n+1) ` ` sum_(r=0)^(n) (C_(r))/(r+ 1)` = `sum_(r=0)^(n) (1)/(r+ 1) . ""^(n)C_(r)` `sum_(r=0)^(n) (1)/(r+ 1) .(n+ 1)/(r+ 1). ""^(n)C_(r)` ` (1)/(r+ 1) sum_(r=0)^(n)(n+ 1)/(r+ 1). ""^(n)C_(r)` ` (1)/(r+ 1) sum_(r=0)^(n). ""^(n)C_(r+1) " "[because ""^(n+1)C_(r +1) = (n+ 1)/(r+ 1). ""^(n)C_(r)]` `= (1) (n+1) {""^(n+1)C_(1)+""^(n+1)C_(2)+""^(n+1)C_(3)+...+""^(n+1)C_(n+1)}` `= (1)/(n+1) (""^(n+1)C_(0)+""^(n+1)C_(1)+...+""^(n+1)C_(n+1))-(""^(n+1)C_(0))}` ` (1)/(n+1) (2^(n+1) -1)` |
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