Question

The coefficient of $x^9$ in the polynomial given by ${\sum }_{t=1}^{11}(x+r)(x+r+1)(x+r+2)........(x+r+9)$ is

• A. $5511$
• B. $5151$
• C. $1515$
• D. $1155$

Answer 1

Answer: (D)

$1155$

$\begin{array}{c}Coeff.of{x}^9{q}_n{\sum }_{r=1}^{11}\left(x+r\right)\left(x+r+1\right)\left(x+r+2\right).....\left(x+r+9\right)\\ ={\sum }_{r=1}^{11}Coff.of{x}^{9}in\left(x+r\right)\left(x+r+1\right)\left(x+r+2\right)+....\left(x+r+9\right)\\ ={\sum }_{r=1}^{11}r+\left(r+1\right)+\left(r+2\right)+\left(r+3\right)+.....+\left(r+9\right)\\ ={\sum }_{r=1}^{11}\left[10r+\left(1+2+\mathrm{3...9}\right)\right]\\ =10{\sum }_{r=1}^{11}r+45{\sum }_{r=1}^{11}.1\\ =10\times \frac{11\times 12}{2}+45\times 11\\ =660+495\\ =1155\\ Hence,theoptionDisthecorrectanswer.\end{array}$