The sum of the coefficients of three consecutive terms in the binomial expansion of , which are in the ratio 1:3:5, is equal to [2023]

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Solution

Q.
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 [2023]

1 41

2 92

3 25

4 63

Ans.
(4)

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

$∴$ $\frac{{}^{n + 2}{C_{r - 1}}}{{}^{n + 2}{C_{r}}} = \frac{1}{3} \Rightarrow n = 4 r - 3 ...(i)$

Also, $\frac{{}^{n + 2}{C_{r}}}{{}^{n + 2}{C_{r + 1}}} = \frac{3}{5} \Rightarrow 8 r - 1 = 3 n ...(ii)$

From (i) and (ii), we get

$4 r - 3 = \frac{8 r - 1}{3} \Rightarrow 4 r = 8 \Rightarrow r = 2 and n = 5$

$Required sum = {}^{7}{C_{1}} + {}^{7}{C_{2}} + {}^{7}{C_{3}} = 7 + 21 + 35 = 63$

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