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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 `xC_(0)-(x -1) C_(1)+(x-2)C_(2)-(x -3)C_(3)+...+(-1)^(n) (x -n) C_(n)=` |
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Answer» Correct Answer - a We have, `xC_(0)-(x -1) C_(1)+(x-2)C_(2)-(x -3)C_(3)+...+(-1)^(n) (x -n) C_(n)=` = `sum_(r=1)^(n) (-1)^(r) (x -r) ""^(n)C_(r)` `x{sum_(x=0)^(n) (-1)^(r) ""^(n)C_(r)} - {sum_(x=0)^(n) (-1)^(r)r ""^(n)C_(r)} ` `x{sum_(x=0)^(n) (-1)^(r) ""^(n)C_(r)} - {sum_(x=0)^(n) (-1)^(r)r.(n)/(r) ""^(n-1)C_(r-1)} ` `x{sum_(x=0)^(n) (-1)^(r) ""^(n)C_(r)} +n {sum_(x=0)^(n) (-1)^(r-1). ""^(n-1)C_(r-1)} ` = `x xx = + n xx 0 = 0` . |
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