The greatest coefficient in the expansion of | vedclass

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

Question

(A) The general term in the expansion of $(1 + x)^{2n + 1}$ is given by $T_{r+1} = {}^{2n+1}C_r x^r$.The coefficients are the binomial coefficients ${

Vedclass · Accepted Answer

$\frac{(2n + 1)!}{n!(n + 1)!}$

Vedclass · Answer

(A) The general term in the expansion of $(1 + x)^{2n + 1}$ is given by $T_{r+1} = {}^{2n+1}C_r x^r$.The coefficients are the binomial coefficients ${}^{2n+1}C_r$ for $r = 0, 1, 2, \dots, 2n+1$.For an expansion $(1+x)^N$,the greatest coefficient is the middle term coefficient. Here $N = 2n + 1$,which is odd.For an odd power $N$,there are two middle terms at $r = \frac{N-1}{2}$ and $r = \frac{N+1}{2}$.Substituting $N = 2n + 1$:$r_1 = \frac{2n+1-1}{2} = n$$r_2 = \frac{2n+1+1}{2} = n+1$The greatest coefficients are ${}^{2n+1}C_n$ and ${}^{2n+1}C_{n+1}$.Calculating ${}^{2n+1}C_n = \frac{(2n+1)!}{n!(2n+1-n)!} = \frac{(2n+1)!}{n!(n+1)!}$.Thus,the greatest coefficient is $\frac{(2n + 1)!}{n!(n + 1)!}$.

The greatest coefficient in the expansion of $(1 + x)^{2n + 1}$ is

Similar Questions

The coefficient of $x^7$ in the expansion of $(1 - x - x^2 + x^3)^6$ is

If the coefficients of $r$ th and $(r+1)$ th terms in the expansion of $(3+7x)^{29}$ are equal,then $r$ is equal to

The number of irrational terms in the expansion of $(5^{1/2} + 7^{1/8})^{1024} + (5^{1/2} - 7^{1/8})^{1024}$ is

The total number of terms in the expansion of $(x + a)^{100} + (x - a)^{100}$ after simplification will be

Let $\alpha > 0, \beta > 0$ be such that $\alpha^{3} + \beta^{2} = 4$. If the maximum value of the term independent of $x$ in the binomial expansion of $(\alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}})^{10}$ is $10k$,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module