Evaluate the expression: frac{1}{81^{n}} - {} | vedclass

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

Question

(C) The given expression is $\frac{1}{81^{n}} \left[ 1 - {}^{2n}C_1(10) + {}^{2n}C_2(10^2) - \dots + (-1)^{2n} {}^{2n}C_{2n}(10^{2n}) \right]$.Using t

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The given expression is $\frac{1}{81^{n}} \left[ 1 - {}^{2n}C_1(10) + {}^{2n}C_2(10^2) - \dots + (-1)^{2n} {}^{2n}C_{2n}(10^{2n}) \right]$.Using the binomial expansion formula $(a - b)^m = \sum_{k=0}^{m} {}^{m}C_k a^{m-k} (-b)^k$,we identify $a = 1$,$b = 10$,and $m = 2n$.The expression inside the bracket is $(1 - 10)^{2n} = (-9)^{2n}$.Substituting this back,we get $\frac{(-9)^{2n}}{81^{n}} = \frac{((-9)^2)^n}{81^n} = \frac{81^n}{81^n} = 1$.

Evaluate the expression: $\frac{1}{81^{n}} - {}^{2n}C_1 \frac{10}{81^{n}} + {}^{2n}C_2 \frac{10^2}{81^{n}} - \dots + \frac{10^{2n}}{81^{n}} = $

Similar Questions

$(\sqrt{2} + 1)^6 - (\sqrt{2} - 1)^6 = $

Find $(x+1)^{6}+(x-1)^{6}$. Hence or otherwise,evaluate $(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}$.

Using the Binomial Theorem,evaluate $(96)^{3}$.

Let $r$ be the remainder when $2021^{2020}$ is divided by $2020^2$. Then $r$ lies between

The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3) \ldots (1+x^{100})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module