Question

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2 x)^n$ be in the ratio $2: 5: 8$. Then the coefficient of the term, which is in the middle of these three terms, is _______

Answer 1

$ t_{r+1}={ }^n C_r(2 x)^r$
$ \Rightarrow \frac{{ }^{ n } C_{r-1}(2)^{r-1}}{{ }^{ n } C_r(2)^{ r }}=\frac{2}{5}$
$ \Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5} $
$ \Rightarrow \frac{r}{n-r+1}=\frac{4}{5} \Rightarrow 5 r=4 n-4 r+4$
$ \Rightarrow 9 r=4(n+1) .....$(1)
$ \Rightarrow \frac{{ }^{ n } C_r(2)^{ r }}{{ }^{ n } C_{r+1}(2)^{r+1}}=\frac{5}{8} $
$ \Rightarrow \frac{\frac{n !}{ r !( n - r ) !}}{\frac{ n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4} $
$ \Rightarrow 4 r+4=5 n-5 r \Rightarrow 5 n-4=9 r \ldots$(2)
From (1) and (2)
$\Rightarrow 4 n+4=5 n-4 \Rightarrow n=8$
(1) $\Rightarrow r =4$
so, coefficient of middle term is
${ }^8 C _4 2^4=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120$