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Given `"^(8)C_(1)x(1-x)^(7)+2*^(8)C_(2)x^(2)(1-x)^(6)+3*^(8)C_(3)x^(3)(1-x)^(5)+...+8*x^(8)=ax+b`, then `a+b` isA. `4`B. `6`C. `8`D. `10` |
Answer» Correct Answer - C `(c )` `.^(n)C_(1)x(1-x)^(n-1)+2*^(n)C_(2)x^(2)(1-x)^(n-2)+3*^(n)C_(3)x^(3)(1-x)^(n-3)+....+n*^(n)C_(n)x^(n)` `=sum_(r=1)^(n)r*^(n)C_(r )x^(r )(1-x)^(n-r)` `=sum_(r=1)^(n)n^(n-1)C_(r-1)x^(r)(1-x)^(n-r)` `=nsum_(r=1)^(n).^(n-1)C_(r-1)x*x^(r-1)(1-x)^((n-1)-(r-1))` `=nxsum_(r=1)^(n).^(n-1)C_(r-1)x^(r-1)(1-x)^((n-1)-(r-1))` `=nx[x+(1-x)]^(n-1)` `=nx` `.^(8)C_(1)x(1-x)^(7)+2*^(8)C_(2)x^(2)(1-x)^(6)+3*^(8)C_(3)x^(3)(1-x)^(5)+....+8.x^(8)` `=8x` |
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