The coefficient of x^9 in the polynomial give | vedclass

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(D) Let $P_r(x) = (x+r)(x+r+1)...(x+r+9)$. This is a polynomial of degree $10$. The coefficient of $x^9$ in $P_r(x)$ is the sum of the constants in ea

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$1155$

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(D) Let $P_r(x) = (x+r)(x+r+1)...(x+r+9)$. This is a polynomial of degree $10$. The coefficient of $x^9$ in $P_r(x)$ is the sum of the constants in each factor,which is $r + (r+1) + (r+2) + ... + (r+9)$. This sum is an arithmetic progression with $10$ terms: $S_r = \frac{10}{2} [2r + (10-1)1] = 5(2r + 9) = 10r + 45$. The coefficient of $x^9$ in the total sum $\sum_{r=1}^{11} P_r(x)$ is $\sum_{r=1}^{11} (10r + 45)$. $= 10 \sum_{r=1}^{11} r + \sum_{r=1}^{11} 45 = 10 	imes \frac{11 	imes 12}{2} + 45 	imes 11$. $= 10 	imes 66 + 495 = 660 + 495 = 1155$.

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

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