The coefficient of x^{53} in the expansion su | vedclass

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

Question

(C) The given expression is of the form $\sum_{m=0}^{n} {^{n}C_m a^{n-m} b^m}$,which is the binomial expansion of $(a+b)^n$.Here,$a = (x-3)$,$b = 2$,a

Vedclass · Accepted Answer

$-^{100}C_{53}$

Vedclass · Answer

(C) The given expression is of the form $\sum_{m=0}^{n} {^{n}C_m a^{n-m} b^m}$,which is the binomial expansion of $(a+b)^n$.Here,$a = (x-3)$,$b = 2$,and $n = 100$.Therefore,the expression simplifies to $((x - 3) + 2)^{100} = (x - 1)^{100}$.We can rewrite this as $(-(1 - x))^{100} = (1 - x)^{100}$.The general term in the expansion of $(1 - x)^{100}$ is $T_{r+1} = {^{100}C_r (1)^{100-r} (-x)^r} = {^{100}C_r (-1)^r x^r}$.To find the coefficient of $x^{53}$,we set $r = 53$.The term is ${^{100}C_{53} (-1)^{53} x^{53}} = -{^{100}C_{53}} x^{53}$.Thus,the coefficient of $x^{53}$ is $-{^{100}C_{53}}$.

The coefficient of $x^{53}$ in the expansion $\sum_{m = 0}^{100} {^{100}C_m (x - 3)^{100 - m} \cdot 2^m}$ is

Similar Questions

The coefficient of $x^{-5}$ in the binomial expansion of ${\left( {\frac{{x + 1}}{{{x^{2/3}} - {x^{1/3}} + 1}} - \frac{{x - 1}}{{x - {x^{1/2}}}}} \right)^{10}}$ where $x \ne 0, 1$,is

In the expansion of ${\left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9}$,the term independent of $x$ is

In the expansion of $(x^2 - 2x)^{10}$,the coefficient of $x^{16}$ is

If the coefficients of the $r^{\text{th}}$,$(r+1)^{\text{th}}$,and $(r+2)^{\text{th}}$ terms in the expansion of $(1+x)^n$ are in the ratio $4:15:42$,then $n-r=$

If the term independent of $x$ in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ is $k,$ then $18 k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module