Question

If the sum of the coefficient in the expansion of $(1+2x{)}^n$ is $6561$, the greatest term in the expansion for $x=\frac{1}{2}$ is

• A. $4^{th}$
• B. $5^{th}$
• C. $6^{th}$
• D. $Noneofthese$

Answer 1

The correct option is B $5^{th}$

It is required that you must know, $\left(1\right)$ binomial theorem

$2)$ greatest term in the binomial theorem

$3)$ application of combination.

Now,

Just put $x=1$, so it will look like $3^n=6561$. So $n=8$

so put $x=\frac{1}{2}$, it will look like $(1+1{)}^8=656$

Greatest term in the binomial expansion is ${(}^n{C}_{n/2})$ when n is even(c for combination)

so here $n=8$

Let $T_r$ is the greatest term of a given expansion

$\therefore T_{r+1}<T_r>T_{r-1}$

1. $\frac{T_{r+1}}{T_r}<1$

$\because T_{r+1}={}^8{C}_r(2x{)}^r,T_r={}^8{C}_{r-1}(2x{)}^{r-1}$

$⟹\frac{{}^8{C}_r(2x{)}^r}{{}^8{C}_{r-1}(2x{)}^{r-1}}<1$

$⟹\frac{{}^8{C}_r\left(2x\right)}{{}^8{C}_{r-1}}<1$

$⟹\frac{\frac{8!}{r!(8-r)!}}{\frac{8!}{(9-r)!(r-1)!}}<1$

$⟹\frac{9-r}{r}<1$

$⟹r>\frac{9}{2}⟹r>4.5$

2. $\frac{T_r}{T_{r-1}}>1$

$⟹\frac{{}^8{C}_{r-1}(2x{)}^{r-1}}{{}^8{C}_{r-2}(2x{)}^{r-2}}>1$

$⟹\frac{\frac{8!}{(9-r)!(r-1)!}}{\frac{8!}{(10-r)!(r-2)!}}>1$

$⟹\frac{10-r}{r-1}>1$

$⟹11>2r⟹r<5.5$

From 1 and 2, $4.5<r>5.5⟹r=5$