# The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of

Question:

The coefiicients of the $(r-1)^{\text {th }}, r^{\text {th }}$ and $(r+1)^{\text {th }}$ terms in the expansion of

$(x+1)^{n}$ are in the ratio $1: 3.5$. Find $n$ and $r$.

Solution:

It is known that $(k+1)^{\text {th }}$ 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, $(r-1)^{\text {th }}$ term in the expansion of $(x+1)^{n}$ is $\mathrm{T}_{r-1}={ }^{n} \mathrm{C}_{r-2}(\mathrm{x})^{n-(r-2)}(1)^{(r-2)}={ }^{n} \mathrm{C}_{r-2} \mathrm{x}^{n-r+2}$

$r^{\text {th }}$ term in the expansion of $(x+1)^{n}$ is $T_{r}={ }^{n} C_{r-1}(x)^{n-(t-1)}(1)^{(r-1)}={ }^{n} C_{r-1} x^{n-r+1}$

$(r+1)^{\mathrm{th}}$ term in the expansion of $(x+1)^{n}$ is $\mathrm{T}_{\mathrm{r}+1}={ }^{n} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{\mathrm{n}-\mathrm{r}}(1)^{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}}$

$(r+1)^{\text {th }}$ term in the expansion of $(x+1)^{n}$ is $T_{r+1}={ }^{n} C_{r}(x)^{n-r}(1)^{r}={ }^{n} C_{r} x^{n-r}$

Therefore, the coefficients of the $(r-1)^{\text {th }}, r^{\text {th }}$, and $(r+1)^{\text {th }}$ terms in the expansion of $(x+1)^{n}$ are ${ }^{n} \mathrm{C}_{r-2},{ }^{n} \mathrm{C}_{r-1}$, and ${ }^{n} \mathrm{Cr}$ respectively. Since these coefficients are in the ratio $1: 3: 5$, we obtain

$\frac{{ }^{n} C_{r-2}}{{ }^{n} C_{r-1}}=\frac{1}{3}$ and $\frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}}=\frac{3}{5}$

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

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

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

$\Rightarrow 3 \mathrm{r}-3=\mathrm{n}-\mathrm{r}+2$

$\Rightarrow \mathrm{n}-4 \mathrm{r}+5=0$ $\ldots(1)$

$\frac{{ }^{n} C_{t-1}}{{ }^{n} C_{r}}=\frac{n !}{(r-1) !(n-r+1)} \times \frac{r !(n-r) !}{n !}=\frac{r(r-1) !(n-r) !}{(r-1) !(n-r+1)(n-r) !}$

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

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

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

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

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

$4 r-12=0$

$\Rightarrow r=3$

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

$n-12+5=0$

$\Rightarrow n=7$

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