The sum of the coefficients of three consecutive terms in the...

Ratio of three consecutive terms in the binomial expansion of ${\left(1 + x\right)}^{n + 2}$ is

$Cr - 1 n + 2 : Cr n + 2 : Cr + 1 n + 2 : : 1: 3: 5$

So,

$⇒ \dfrac{Cr n + 2}{Cr - 1 n + 2} = \dfrac{3}{1}$

$⇒ \dfrac{n - r + 3}{r} = 3$

$⇒ n + 3 = 4r . . . \left(1\right)$

And,

$\dfrac{Cr + 1 n + 2}{Cr n + 2} = \dfrac{5}{3}$

$⇒ \dfrac{n + 2 - r}{r + 1} = \dfrac{5}{3}$

$⇒ 3n + 6 - 3r = 5r + 5$

$⇒ 3n + 1 = 8r . . . \left(2\right)$

So, $3\left(4r - 3\right) + 1 = 8r$

$⇒ 12r - 9 + 1 = 8r$

$⇒ 4r = 8$

$⇒ r = 2$

So, $n = 5$

Hence, we have

${\left(1 + x\right)}^{n + 2} = {\left(1 + x\right)}^{7}$

Sum of consecutive terms is

$C1 7 + C2 7 + C3 7$

$= 7 + 21 + 35 = 63$