Question

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x{)}^n$ 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

Answer: 1120

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