Question

The coefficients of three consecutive terms in the expansion of $(1+a{)}^n$ are in the ratio $1:7:42.$Find $n$.

Answer 1

Step - $1$: Simplify given data
Let three consecutive terms be $(r-1{)}^{th},r^{th}$ and $(r+1{)}^{th}$ terms.
$T_{r-1},T_r$ and $T_{r+1}$
We know that general term of expansion $(a+b{)}^n$ is $T_{r+1}={}^n{C}_r\left(a{)}^{n-r}\right(b)r$
Ratio of three consecutive terms
$T_{r-2+1}:T_{r-1+1}:T_{r+1}$
$\Rightarrow {}^n{C}_{r-2}\left(1{)}^{n-r+2}\right(a{)}^{r-2}:{}^n{C}_{r-1}\left(1{)}^{n-r+1}\right(a{)}^{r-1}:{}^n{C}_r\left(1{)}^{n-r}\right(a{)}^r$
$\Rightarrow {}^n{C}_{r-2}(a{)}^{r-2}:{}^n{C}_{r-1}(a{)}^{r-1}:{}^n{C}_r(a{)}^r$
Ratio of coefficient of three consecutive terms
$\Rightarrow {}^n{C}_{r-2}:{}^n{C}_{r-1}:{}^n{C}_r$

Step $2$: Solve equation for value of $n$
Since the coefficient of $(r-1{)}^{th},r^{th}$ and $(r+1{)}^{th}$ terms are in ration $1:7:42$
Now. ${}^n{C}_{r-2}:{}^n{C}_{r-1}:{}^n{C}_r=1:7:42$
Comparing
$\frac{{}^n{C}_{r-2}}{{}^n{C}_{r-1}}=\frac{1}{7}...\left(1\right)$
and $\frac{{}^n{C}_{r-2}}{{}^n{C}_r}=\frac{7}{42}...\left(2\right)$
From equation $\left(1\right)$
$\frac{{}^n{C}_{r-2}}{{}^n{C}_{r-1}}=\frac{1}{7}$
$\Rightarrow \frac{n!}{(r-2)![n-r+2]!}\times \frac{(r-1)![n-r+1]!}{n!}=\frac{1}{7}$
$\Rightarrow \frac{(r-1)\times (r-2)!\times [n-r+1]!}{(r-2)![n-r+2]\times [n-r+1]!}=\frac{1}{7}\Rightarrow \frac{r-1}{n-r+2}=\frac{1}{7}$
$\Rightarrow 7r-7=n-r+2\Rightarrow n-8r+9=0...\left(3\right)$

From equation $\left(2\right)$
$\Rightarrow \frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{7}{42}$
$\Rightarrow \frac{n!}{(r-1)!(n-r+1)!}\times \frac{r!(n-r)!}{n!}=\frac{1}{6}$

$\Rightarrow \frac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)(n-r)!}=\frac{1}{6}$

$\Rightarrow \frac{r}{n-r+1}=\frac{1}{6}$
$\Rightarrow 6r=n-r+1$
$\Rightarrow n-7r+1=0...\left(4\right)$

Equation $\left(4\right)-$ equation $\left(3\right)$, we get
$\Rightarrow n-7r+1-n+8r-9=0$
$\Rightarrow r-8=0$
$\Rightarrow r=8$
Putting value of $r$ in equation $\left(4\right)$
$\Rightarrow n-7\left(8\right)+1=0$
$\Rightarrow n-56+1-0$
$\Rightarrow n=55$