Let the ratio of the fifth term from the begi | vedclass

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

Question

(A) The binomial expansion is $(\sqrt[4]{2} + 3^{-1/4})^n$. Let $T_r$ denote the $r$-th term from the beginning. The $5$-th term from the beginning is

Vedclass · Accepted Answer

$84$

Vedclass · Answer

(A) The binomial expansion is $(\sqrt[4]{2} + 3^{-1/4})^n$. Let $T_r$ denote the $r$-th term from the beginning. The $5$-th term from the beginning is $T_5 = {^nC_4} (2^{1/4})^{n-4} (3^{-1/4})^4$. The $5$-th term from the end is the $(n-5+2) = (n-3)$-th term from the beginning,which is $T_{n-3} = {^nC_{n-4}} (2^{1/4})^4 (3^{-1/4})^{n-4} = {^nC_4} (2^{1/4})^4 (3^{-1/4})^{n-4}$. The ratio $\frac{T_5}{T_{n-3}} = \frac{(2^{1/4})^{n-4} (3^{-1/4})^4}{(2^{1/4})^4 (3^{-1/4})^{n-4}} = \frac{2^{(n-8)/4}}{3^{(8-n)/4}} = 2^{(n-8)/4} \cdot 3^{(n-8)/4} = 6^{(n-8)/4}$. Given $6^{(n-8)/4} = 6^{1/4}$,so $n-8 = 1$,which gives $n = 9$. The $6$-th term from the beginning is $T_6 = {^9C_5} (2^{1/4})^4 (3^{-1/4})^5 = 126 \cdot 2 \cdot 3^{-5/4} = \frac{252}{3 \cdot 3^{1/4}} = \frac{84}{3^{1/4}}$. Thus,$\alpha = 84$.

Similar Questions

The coefficient of ${x^{-9}}$ in the expansion of ${\left( \frac{x^2}{2} - \frac{2}{x} \right)^9}$ is

Show that the middle term in the expansion of $(1+x)^{2n}$ is $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} 2^n x^n$,where $n$ is a positive integer.

If the coefficients of the $(2r+6)^{\text{th}}$ and $(r-1)^{\text{th}}$ terms in the expansion of $(1+x)^{21}$ are equal,then the value of $r$ is:

Let $[t]$ denote the greatest integer $\leq t$. If the constant term in the expansion of $\left(3x^2 - \frac{1}{2x^5}\right)^7$ is $\alpha$,then $[\alpha]$ is equal to $............$.

If the eleventh term in the binomial expansion of $(x+a)^{15}$ is the geometric mean of the eighth and twelfth terms,then the greatest term in the expansion is

Vedclass Test Series

Exam Paper Generator

Online Exam Module