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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

Which is larger: $99^{50} + 100^{50}$ or $101^{50}$?

Prove that $\sum\limits_{r = 0}^n {3^r \,^nC_r} = 4^n$.

For $|x| < \frac{1}{5}$,the coefficient of $x^3$ in the expansion of $\frac{1}{(1-5 x)(1-4 x)}$ is

Let $[x]$ denote the greatest integer less than or equal to $x$. If for $n \in N$,$(1-x+x^3)^n = \sum_{j=0}^{3n} a_j x^j$,then $\sum_{j=0}^{[\frac{3n}{2}]} a_{2j} + 4 \sum_{j=0}^{[\frac{3n-1}{2}]} a_{2j+1}$ is equal to

If $b$ is very small as compared to the value of $a$,so that the cube and other higher powers of $\frac{b}{a}$ can be neglected in the identity $\frac{1}{a-b}+\frac{1}{a-2b}+\frac{1}{a-3b}+\ldots+\frac{1}{a-nb}=\alpha n+\beta n^2+\gamma n^3$,then the value of $\gamma$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module