Let the coefficients of three consecutive terms in the binomial...

Given,

The coefficients of three consecutive terms in the binomial expansion of $( 1 + 2x{)}^{n}$ be in the ratio $2: 5: 8$.

Now ${r}^{th}$ term in expansion of $( 1 + 2x{)}^{n} is given by,${t}_{r + 1} = Cr n ( 2x{)}^{r}$</p><p>Now let${T}_{r}, {T}_{r + 1}, {T}_{r + 2}$are in the ratio$2: 5: 8$</p><p>$⇒ \dfrac{{T}_{r}}{{T}_{r + 1}} = \dfrac{Cr - 1 n ( 2{)}^{r - 1}}{Cr n ( 2{)}^{r}} = \dfrac{2}{5}$</p><p>$⇒ \dfrac{\dfrac{n !}{( r - 1 ) ! ( n - r + 1 ) !}}{\dfrac{n ! ( 2 )}{r ! ( n - r ) !}} = \dfrac{2}{5}$</p><p>$⇒ \dfrac{r}{n - r + 1} = \dfrac{4}{5} ⇒ 5r = 4n - 4r + 4$</p><p>$⇒ 9r = 4 ( n + 1 ) . . . . . . . \left(1\right)$</p><p>Now taking other ratio$\dfrac{{T}_{r + 1}}{{T}_{r + 2}} = \dfrac{Cr n ( 2{)}^{r}}{Cr + 1 n ( 2{)}^{r + 1}} = \dfrac{5}{8}$</p><p>$⇒ \dfrac{\dfrac{n !}{r ! ( n - r ) !}}{\dfrac{n !}{( r + 1 ) ! ( n - r - 1 ) !}} = \dfrac{5}{4} ⇒ \dfrac{r + 1}{n - r} = \dfrac{5}{4}$</p><p>$⇒ 4r + 4 = 5n - 5r$</p><p>$⇒ 5n - 4 = 9r . . . . . . . . . \left(2\right)$</p><p>From equation \left(1\right) \text{ and } \left(2\right)$ we get,

$4n + 4 = 5n - 4 ⇒ n = 8$ and $r = 4$

So, coefficient of middle term will be

$C4 8 {2}^{4} = 16 \times \dfrac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 16 \times 70 = 1120$