If the number of integral terms in the expans | vedclass

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

Question

(B) The general term of the expansion is given by $T_{r+1} = {^nC_r} (3)^{\frac{n-r}{2}} (5)^{\frac{r}{8}}$,where $0 \le r \le n$.For the term to be i

Vedclass · Accepted Answer

$256$

Vedclass · Answer

(B) The general term of the expansion is given by $T_{r+1} = {^nC_r} (3)^{\frac{n-r}{2}} (5)^{\frac{r}{8}}$,where $0 \le r \le n$.For the term to be integral,both exponents $\frac{n-r}{2}$ and $\frac{r}{8}$ must be non-negative integers.This implies $r$ must be a multiple of $8$,i.e.,$r \in \{0, 8, 16, \dots, 8k\}$.Also,$\frac{n-r}{2}$ must be an integer,which means $(n-r)$ must be even. Since $r$ is a multiple of $8$ (even),$n$ must also be even.Given there are exactly $33$ integral terms,the possible values of $r$ are $0, 8, 16, \dots, 8 	imes 32$.The largest value of $r$ is $8 	imes 32 = 256$.Since $r \le n$,the smallest possible value for $n$ such that $r=256$ is allowed is $n = 256$ (as $n$ must be even and $n \ge 256$ to accommodate $33$ terms).Thus,the least value of $n$ is $256$.

If the number of integral terms in the expansion of $(3^{\frac{1}{2}} + 5^{\frac{1}{8}})^n$ is exactly $33$,then the least value of $n$ is

Similar Questions

The coefficient of $x^2$ in the expansion of $(2x^2 + \frac{1}{x})^{10}, x \neq 0$,is:

The coefficient of $x^n$ in the expansion of $(1 + x)(1 - x)^n$ is

If the $9^{th}$ and $10^{th}$ terms are the numerically greatest terms in the expansion of $(5x - 6y)^n$ when $x = 2/5$ and $y = 1/2$,then the absolute value of the middle term of that expansion is:

The $16^{th}$ term in the expansion of $(\sqrt{x} - \sqrt{y})^{17}$ is

If the third term in the binomial expansion of $(1 + x^{\log_2 x})^5$ equals $2560$,then a possible value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module