The coefficient of x^{7} in the expression (1 | vedclass

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

Question

(B) The given expression is a geometric series with first term $a = (1+x)^{10}$,common ratio $r = \frac{x}{1+x}$,and $n = 11$ terms.Sum $S = a \frac{1

Vedclass · Accepted Answer

$330$

Vedclass · Answer

(B) The given expression is a geometric series with first term $a = (1+x)^{10}$,common ratio $r = \frac{x}{1+x}$,and $n = 11$ terms.Sum $S = a \frac{1-r^{n}}{1-r} = (1+x)^{10} \frac{1-(\frac{x}{1+x})^{11}}{1-\frac{x}{1+x}} = (1+x)^{10} \frac{1-\frac{x^{11}}{(1+x)^{11}}}{\frac{1+x-x}{1+x}} = (1+x)^{11} \left(1-\frac{x^{11}}{(1+x)^{11}}ight) = (1+x)^{11}-x^{11}$.We need the coefficient of $x^{7}$ in $(1+x)^{11}-x^{11}$.The general term in the expansion of $(1+x)^{11}$ is given by $^{11}C_{r} x^{r}$.For $r = 7$,the coefficient is $^{11}C_{7} = \frac{11!}{7!4!} = \frac{11 	imes 10 	imes 9 	imes 8}{4 	imes 3 	imes 2 	imes 1} = 330$.

The coefficient of $x^{7}$ in the expression $(1+x)^{10}+x(1+x)^{9}+x^{2}(1+x)^{8}+\ldots+x^{10}$ is

Similar Questions

The least value of $n$ for which the number of integral terms in the Binomial expansion of $(\sqrt[3]{7}+\sqrt[12]{11})^{n}$ is $183$ is:

If the coefficient of $x^{8}$ in $\left(a x^{2}+\frac{1}{b x}\right)^{13}$ is equal to the coefficient of $x^{-8}$ in $\left(a x-\frac{1}{b x^{2}}\right)^{13},$ then $a$ and $b$ will satisfy the relation

If the coefficients of the $(2r + 1)^{th}$ term and the $(r + 2)^{th}$ term are equal in the expansion of $(1 + x)^{43}$,then the value of $r$ is:

The coefficient of $x^4$ in the expansion of $(1-x+x^2-x^3)^4$ is

If the $4^{\text{th}}$ term in the expansion of $\left(\frac{x}{2}-\frac{2y}{3}\right)^6$ is $-20$,then $xy=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module