The coefficient of x^8 in the expansion of (x | vedclass

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(B) Let $P(x) = (x-a_1)(x-a_2)\ldots(x-a_n) = x^n - S_1 x^{n-1} + S_2 x^{n-2} - \ldots + (-1)^n S_n$,where $S_k$ is the sum of products of $a_1, a_2,

Vedclass · Accepted Answer

$1320$

Vedclass · Answer

(B) Let $P(x) = (x-a_1)(x-a_2)\ldots(x-a_n) = x^n - S_1 x^{n-1} + S_2 x^{n-2} - \ldots + (-1)^n S_n$,where $S_k$ is the sum of products of $a_1, a_2, \ldots, a_n$ taken $k$ at a time.Here,$n=10$ and $a_i = i$ for $i=1, 2, \ldots, 10$.The coefficient of $x^8$ is $S_2$,which is the sum of products of $1, 2, \ldots, 10$ taken $2$ at a time.$S_2 = \sum_{1 \le i We know $\sum_{i=1}^{10} i = \frac{10(11)}{2} = 55$.We know $\sum_{i=1}^{10} i^2 = \frac{10(11)(21)}{6} = 385$.$S_2 = \frac{1}{2} [55^2 - 385] = \frac{1}{2} [3025 - 385] = \frac{1}{2} [2640] = 1320$.

The coefficient of $x^8$ in the expansion of $(x-1)(x-2)(x-3)\ldots(x-10)$ is:

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