If the coefficients of three consecutive terms in the expansion of (1+x)^n are in the rati…

Question

If the coefficients of three consecutive terms in the expansion of $(1+x)^{n}$ are in the ratio $1: 5: 20$, then the coefficient of the fourth term is

Accepted Answer

$$
\begin{aligned}
& \text { Given: }{ }^n \mathrm{C}_{r-1}:{ }^n \mathrm{C}_r:{ }^n \mathrm{C}_{r+1} \\
& =1: 5: 20 \\
& \Rightarrow \frac{n !}{(r-1) !(n-r+1) !} \times \frac{r !(n-r) !}{n !}=\frac{1}{5} \\
& \Rightarrow \frac{r}{(n-r+1)}=\frac{1}{5} \\
& \Rightarrow 5 r=n-r+1 \\
& \Rightarrow n=6 r-1 ........(i)
\end{aligned}
$$

$$
\begin{aligned}
& \text { Also, } \frac{n}{r !(n-r) !} \times \frac{(r+1) !(n-r-1) !}{n !}=\frac{5}{20}=\frac{1}{20} \\
& \Rightarrow \frac{(r+1)}{(n-r)}=\frac{1}{4} \\
& \Rightarrow 4 r+4=n-r \\
& \Rightarrow n=5 r+4 ..........(ii)
\end{aligned}
$$

From (i) and (ii), we get

$$
\begin{aligned}
& 6 r-1=5 r+4 \\
& \Rightarrow r=5 \\
& \text { So, } n=5(5)+4=29 \\
& \text { So, coefficient of } 4{ }^{\text {th }} \text { terms }={ }^n \mathrm{C}_3={ }^{29} \mathrm{C}_3 \\
& =\frac{29 !}{3 ! 26 !}=\frac{29 \times 28 \times 27}{3 \times 2}=3654
\end{aligned}
$$

Solutions · 1