If the second term of the expansion left[ a^{ | vedclass

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

Question

(A) The given expansion is $\left[ a^{1/13} + a \cdot (a^{-1})^{-1/2} \right]^n = \left[ a^{1/13} + a \cdot a^{1/2} \right]^n = \left[ a^{1/13} + a^{3

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The given expansion is $\left[ a^{1/13} + a \cdot (a^{-1})^{-1/2} ight]^n = \left[ a^{1/13} + a \cdot a^{1/2} ight]^n = \left[ a^{1/13} + a^{3/2} ight]^n$.The general term $T_{r+1}$ is given by $^nC_r (a^{1/13})^{n-r} (a^{3/2})^r$.The second term $(r=1)$ is $T_2 = ^nC_1 (a^{1/13})^{n-1} (a^{3/2})^1 = n \cdot a^{\frac{n-1}{13}} \cdot a^{3/2} = n \cdot a^{\frac{n-1}{13} + \frac{3}{2}}$.Given $T_2 = 14a^{5/2}$,we have $n \cdot a^{\frac{n-1}{13} + \frac{3}{2}} = 14a^{5/2}$.Comparing the coefficients,$n = 14$.Checking the exponent: $\frac{14-1}{13} + \frac{3}{2} = 1 + 1.5 = 2.5 = 5/2$,which matches.Now,calculate $\frac{^nC_3}{^nC_2} = \frac{^{14}C_3}{^{14}C_2} = \frac{\frac{14 	imes 13 	imes 12}{3 	imes 2 	imes 1}}{\frac{14 	imes 13}{2 	imes 1}} = \frac{12}{3} = 4$.

If the second term of the expansion $\left[ a^{\frac{1}{13}} + \frac{a}{\sqrt{a^{-1}}} \right]^n$ is $14a^{5/2}$,then the value of $\frac{^nC_3}{^nC_2}$ is:

Similar Questions

Let $a_n$ denote the term independent of $x$ in the expansion of $\left[x+\frac{\sin(1/n)}{x^2}\right]^{3n}$. Then $\lim_{n \to \infty} \frac{a_n \cdot n!}{^{3n}P_n}$ equals

If the $m$-th term is the middle term in the expansion of $\left(x^2-\frac{1}{2x}\right)^{20}$,find the coefficient of $T_{m+3}$.

If the $1011^{\text{th}}$ term from the end in the binomial expansion of $(\frac{4x}{5} - \frac{5}{2x})^{2022}$ is $1024$ times the $1011^{\text{th}}$ term from the beginning,then $|x|$ is equal to

The coefficient of $\frac{1}{x}$ in the expansion of $(1 + x)^n (1 + \frac{1}{x})^n$ 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

Vedclass Test Series

Exam Paper Generator

Online Exam Module