binom{n}{0} + 2binom{n}{1} + 2^2binom{n}{2} + | vedclass

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

Question

(C) We know the binomial expansion: $(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k = \binom{n}{0} + x\binom{n}{1} + x^2\binom{n}{2} + \dots + x^n\binom{

Vedclass · Accepted Answer

$3^n$

Vedclass · Answer

(C) We know the binomial expansion: $(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k = \binom{n}{0} + x\binom{n}{1} + x^2\binom{n}{2} + \dots + x^n\binom{n}{n}$.By substituting $x = 2$ into the expansion,we get:$(1 + 2)^n = \binom{n}{0} + 2\binom{n}{1} + 2^2\binom{n}{2} + \dots + 2^n\binom{n}{n}$.Therefore,the given expression is equal to $3^n$.

$\binom{n}{0} + 2\binom{n}{1} + 2^2\binom{n}{2} + \dots + 2^n\binom{n}{n}$ is equal to

Similar Questions

The sum of the coefficients of the terms of degree $m$ in the expansion of $(1 + x)^n(1 + y)^n(1 + z)^n$ is:

The approximate value of $(1.0002)^{3000}$ is

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

The roots of the equation $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ are

For $|x| < 1$,the constant term in the expansion of $\frac{1}{(x-1)^2(x-2)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module