Question

If the coefficients of three consecutive terms in the expansion of ${(1+x)}^n$ are in the ratio $1:6:30,$ then the number of terms in the expansion is

Answer 1

Let the coefficients of $(r+1{)}^{\text{th}},(r+2{)}^{\text{th}},(r+3{)}^{\text{th}}$ be in the ratio $1:6:30$

$\frac{{}^n{C}_r}{{}^n{C}_{r+1}}=\frac{1}{6}\Rightarrow \frac{r+1}{n-r}=\frac{1}{6}\Rightarrow n-7r=6\cdots \left(1\right)$

$\frac{{}^n{C}_{r+1}}{{}^n{C}_{r+2}}=\frac{6}{30}\Rightarrow \frac{r+2}{n-r-1}=\frac{1}{5}\Rightarrow n-6r=11\cdots \left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right)$, we get
$r=5$ and $n=41$

Hence, the number of terms in the expansion is $n+1=42$