QJEEM23Ma337f: JEE Main 2023 Mathematics - Class 11 JEEM23Ma337f Solution: The sum of the coefficients of three consecutive terms in the…

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.

$\begin{aligned}& \frac{T_{r+1}}{T_r}=\frac{{ }^{n+2} C_r}{{ }^{n+2} C_{r-1}}=\frac{n+2-r+1}{r}=\frac{n+3-r}{r}=3 \\\\& n-4 r+3=0 ......(1) \\\\& \frac{T_{r+2}}{T_{r+1}}=\frac{{ }^{n+2} C_{r+1}}{{ }^{n+2} C_r}=\frac{(n+2)-(r+1)+1}{r+1}=\frac{n-r+2}{r+1}=\frac{5}{3} \\\\& 3 n-8 r+1=0 ......(2)\end{aligned}$

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

$\Rightarrow n=5, r=2$

$\begin{aligned}T_r+T_{r+1}+T_{r+2} & ={ }^7 C_1+{ }^7 C_2+{ }^7 C_3 \\\\& =7+21+35=63\end{aligned}$