Question

The coefficient of the ${\left(r-1\right)}^{th},r^{th}and{\left(r+1\right)}^{th}$ terms in the expansion of ${\left(x+1\right)}^n$ are in the ratio $1:3:5.$ Find $n$ and $r$

Answer 1

It is known that $(k+1{)}^{th}$ term, $\left(T_{k+1}\right)$, in the binomial expansion of $(a+b{)}^n$ is given by

$T_{k+1}={}^n{C}_k{a}^{n-k}{b}^k$.

Therefore, $(r-1{)}^{th}$ term in the expansion of $(x+1{)}^n$ is

$T_{r-1}={}^n{C}_{r-2}\left(x{)}^{n-(r-2)}\right(1){=}^{(r-2)}={}^n{C}_{r-2}{x}^{n-r+2}$

$(r+1)$ term in the expansion of $(x+1{)}^n$ is

$T_{r+1}={}^n{C}_r\left(x{)}^{n-r}\right(1{)}^r={}^n{C}_r{x}^{n-r}$

$r^{th}$ term in the expansion of $(x+1{)}^n$ is

$T_r{=}^n{C}_{r-1}\left(x{)}^{n-(r-1)}\right(1{)}^{(r-1)}={}^n{C}_{r-1}{x}^{n-r+1}$

Therefore, the coefficients of the $(r-1{)}^{th},r^{rh}$ and $(r+1{)}^{th}$ terms in the expansion of $(x+1{)}^n$

${}^n{C}_{r-2},{}^n{C}_{r-1}$, and ${}^n{C}_r$, are respectively. Since these coefficients are in the ratio $1:3:5$, we obatin

$\frac{{}^n{C}_{r-2}}{{}^n{C}_{r-1}}=\frac{1}{3}$ and $\frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{3}{5}$

$\frac{{}^n{C}_{r-2}}{{}^n{C}_{r-1}}=\frac{n!}{(r-2)!(n-r+1)!}\times \frac{(r-1)!(n-r+1)!}{n!}=\frac{(r-1)(r-2)!(n-r+1)!}{(r-2)!(n-r+2)!(n-r+1)!}$

$=\frac{r-1}{n-r+2}$

$\therefore \frac{r-1}{n-r+2}=\frac{1}{3}$

$\Rightarrow 3r-3=n-r+2$

$\Rightarrow n-4r+5=0$ ......(1)

$\frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{n!}{(r-1)!(n-r+1)}\times \frac{r!(n-r)!}{n!}=\frac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)(n-r)!}$

$=\frac{r}{n-r+1}$

$\therefore \frac{r}{n-r+1}=\frac{3}{5}$

$\to 5r=3n-3r+3$

$\Rightarrow 3n-8r+3=0$ ..(2)

Multiplying (1) by 3 and subtracting it from (2), we obtain

$4r-12=0$

$\Rightarrow r=3$

Putting the value of $r$ in (1), we obtain $n$

$-12+5=0$

$\Rightarrow n=7$

Thus, $n=7$ and $r=3$