how we will suppose d 3 consecutive terms in expansion of (1+a) raise to power n if d coefficients of - Maths - Binomial Theorem

Question

how we will suppose d 3 consecutive terms in expansion of (1+a)
raise to power n if d coefficients of these terms r in ratio
1;7;42?also find n?

Santosh Kumar · Accepted Answer

Let the three consecutive terms be rth, (r + 1)th and (r + 2)th terms In the expansion of (1 + x)n , coefficients of rth, (r + 1)th and (r + 2)th terms are Cr-1n, Crn and Cr+1n respectively ${}^{n}{C_{r - 1}} : {}^{n}{C_{r}} : {}^{n}{C_{r + 1}} = 1 : 7 : 42 \frac{{}^{n}{C_{r - 1}}}{{}^{n}{C_{r}}} = \frac{1}{7} \frac{n!}{(r - 1)! (n - r + 1)!} \times \frac{(n - r)! r!}{n!} = \frac{1}{7} \Rightarrow \frac{(n - r)! r (r - 1)!}{(r - 1)! (n - r + 1) (n - r)!} = \frac{1}{7} \Rightarrow \frac{r}{n - r + 1} = \frac{1}{7} \Rightarrow 7 r = n - r + 1 \Rightarrow 7 r + r = n + 1 \Rightarrow 8 r = n + 1 \Rightarrow n = 8 r - 1 (1) \Rightarrow \frac{{}^{n}{C_{r}}}{{}^{n}{C_{r + 1}}} = \frac{7}{42}$ $\frac{n!}{r! (n - r)!} \times \frac{(r + 1)! (n - r - 1)!}{n!} = \frac{7}{42} \Rightarrow \frac{n! (r + 1) r! (n - r - 1)!}{r! (n - r) (n - r - 1)! n!} = \frac{7}{42} \Rightarrow \frac{r + 1}{n - r} = \frac{1}{6} \Rightarrow 6 r + 6 = n - r \Rightarrow 6 r + 6 = 8 r - 1 - r [using equation (1)] \Rightarrow 7 r - 1 = 6 r + 6 \Rightarrow 7 r - 6 r = 6 + 1 \Rightarrow r = 7 ∴ n = 8 (7) - 1 = 55$

how we will suppose d 3 consecutive terms in expansion of (1+a) raise to power n if d coefficients of these terms r in ratio 1;7;42?also find n?