The sum of the coefficients of three consecutive terms in the binomial expansion of [(1+x)…

Question

The sum of the coefficients of three consecutive terms in the binomial expansion of $$(1+\mathrm{x})^{\mathrm{n}+2}$$, which are in the ratio $1: 3: 5$, is equal to :

Accepted Answer

The problem asks for 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.

Given that the ratios of the coefficients are 1:3:5, we let the terms be ${T}_{r}$, ${T}_{r+1}$, and ${T}_{r+2}$. The coefficients of these terms are ${}^{n+2}{C}_{r-1}$, ${}^{n+2}{C}_{r}$, and ${}^{n+2}{C}_{r+1}$, respectively.

$\begin{gathered} \frac{{T}_{r+1}}{{T}_{r}}=\frac{{}^{n+2}{C}_{r}}{{}^{n+2}{C}_{r-1}}=\frac{n+2-r+1}{r}=\frac{n+3-r}{r}=3 \ n-4r+3=0......(1) \ \frac{{T}_{r+2}}{{T}_{r+1}}=\frac{{}^{n+2}{C}_{r+1}}{{}^{n+2}{C}_{r}}=\frac{(n+2)-(r+1)+1}{r+1}=\frac{n-r+2}{r+1}=\frac{5}{3} \ 3n-8r+1=0......(2) \end{gathered}$

By solving (1) and (2), we get

$\Rightarrow n=5,r=2$

$\begin{gathered} {T}_{r}+{T}_{r+1}+{T}_{r+2} ={}^{7}{C}_{1}+{}^{7}{C}_{2}+{}^{7}{C}_{3} \ \ =7+21+35=63 \end{gathered}$

Solutions · 1