Question

If the coefficient of $(r-1{)}^{th},r^{th}$ and $(r+1{)}^{th}$ terms in the expansion of $(x+1{)}^n$ are in the ratio $1:3:5$, then

• A. $r=3$
• B. $r=4$
• C. $n=7$
• D. $n=8$

Answer 1

Answer: (A), (C)

$n=7$

$(x+1{)}^n=(1+x{)}^n$

$T_{r-1}=T_{(r-2)+1}={}^n{C}_{r-2}{x}^{r-2}$
$T_r=T_{(r-1)+1}={}^n{C}_{r-1}{x}^{r-1}$
$T_{r+1}={}^n{C}_r{x}^r$

Given that,
coeff. of $T_{r-1}:$ coeff. of $T_r:$ coeff. of $T_{r+1}=1:3:5$
$\Rightarrow {}^n{C}_{r-2}:{}^n{C}_{r-1}:{}^n{C}_r=1:3:5\Rightarrow \frac{{}^n{C}_{r-2}}{{}^n{C}_{r-1}}=\frac{1}{3}\text{and}\frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{3}{5}$

$\frac{{}^n{C}_{r-2}}{{}^n{C}_{r-1}}=\frac{1}{3}\Rightarrow \frac{n!}{(r-2)!\cdot (n-r+2)!}\times \frac{(r-1)!(n-r+1)!}{n!}=\frac{1}{3}\Rightarrow \frac{r-1}{n-r+2}=\frac{1}{3}\Rightarrow n-4r=-5\cdots \left(1\right)$

Also, $\frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{3}{5}$ $\Rightarrow \frac{r}{n-r+1}=\frac{3}{5}\Rightarrow 3n-8r=-3\cdots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right),$ we get
$n=7$ and $r=3$