The sum of the coefficients of three consecutive terms in the binomial expansion of 1 + x n + 2 , which are in the ratio 1 : 3 : 5 , is equal to

Question

Aicharya · Accepted Answer

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

${}^{n + 2}C_{r - 1} : {}^{n + 2}C_{r} : {}^{n + 2}C_{r + 1} : : 1 : 3 : 5$

So,

$\Rightarrow \frac{{}^{n + 2}C_{r}}{{}^{n + 2}C_{r - 1}} = \frac{3}{1}$

$\Rightarrow \frac{n - r + 3}{r} = 3$

$\Rightarrow n + 3 = 4 r . . . (1)$

And,

$\frac{{}^{n + 2}C_{r + 1}}{{}^{n + 2}C_{r}} = \frac{5}{3}$

$\Rightarrow \frac{n + 2 - r}{r + 1} = \frac{5}{3}$

$\Rightarrow 3 n + 6 - 3 r = 5 r + 5$

$\Rightarrow 3 n + 1 = 8 r . . . (2)$

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

$\Rightarrow 12 r - 9 + 1 = 8 r$

$\Rightarrow 4 r = 8$

$\Rightarrow r = 2$

So, $n = 5$

Hence, we have

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

Sum of consecutive terms is

${}^{7}C_{1} + {}^{7}C_{2} + {}^{7}C_{3}$

$= 7 + 21 + 35 = 63$

The sum of the coefficients of three consecutive terms in the binomial expansion of 1 + x n + 2 , which are…