InterviewSolution
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InterviewSolution
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If `(1+x)^(n) = C_(0)+C_(1)x + C_(2) x^(2) +...+C_(n)x^(n)` then ` C_(0)""^(2)+C_(1)""^(2) + C_(2)""^(2) +...+C_(n)""^(2)` is equal toA. `2^(2n-2)`B. `2^(n)`C. `((2n)!)/(2(2!)1^(2))`D. `((2n)!)/((n!)^(2))` |
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Answer» Correct Answer - d Using binomial expansion, we have `(1+x)^(n) = C_(0)+C_(1)x + C_(2) x^(2) +...+C_(r)x^(n)+...+C_(n)x^(n)" " ...(i)` `(1+x)^(n) = C_(0)x^(n)+C_(1)x^(n-1) + C_(2) x^(n-2) +...+C_(r)x^(n-r)+...+C_(n-1)x+C_(n)" " ...(ii)` Multiplying (i) and (ii), we get `(1 + x)^(2n) = ( C_(0)+C_(1)x + C_(2)x""^(2) +...+C_(r)x""^(r) + ...+ C_(n)x^(n))` `xx ( C_(0)x^(n)+C_(1)x^(n-1) + C_(2)x""^(n-2) +...+C_(n-r)x""^(r) + ...+ C_(n-1)x+C_(n))` or, `(C_(0) + C_(1) x + C_(2) x^(2) + ...+C_(r)n^(n-2)+...+C_(r) x^(n-r) + ...+C_(n)x^(n))` `xx( C_(0)x^(n)+C_(1)x^(n-1) + C_(2) x^(n-2) +...+C_(r)x^(n-r)+...+C_1)x+C_(n)` `(1 + x)^(2n)`" " ...(iii) Equating the coefficients of `x^(n)` on both sides of (iii) , we get `C_(0)^(2)+C_(1)^(2)+C_(2)^(2)+...+ C_(n)^(2)= ""^(2n)C_(n)` `rArr C_(0)^(2)+C_(1)^(2)+C_(2)^(2)+...+ C_(n)^(2)= ((2n)!)/(n!n!)` . |
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