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 :

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. <br/><br/>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. <br/><br/> $$ \begin{aligned} & \frac{T_{r+1}}{T_r}=\frac{{ }^{n+2} C_r}{{ }^{n+2} C_{r-1}}=\frac{n+2-r+1}{r}=\frac{n+3-r}{r}=3 \\\\ & n-4 r+3=0 ......(1) \\\\ & \frac{T_{r+2}}{T_{r+1}}=\frac{{ }^{n+2} C_{r+1}}{{ }^{n+2} C_r}=\frac{(n+2)-(r+1)+1}{r+1}=\frac{n-r+2}{r+1}=\frac{5}{3} \\\\ & 3 n-8 r+1=0 ......(2) \end{aligned} $$ <br/><br/>By solving (1) and (2), we get <br/><br/>$\Rightarrow n=5, r=2$ <br/><br/>$$ \begin{aligned} T_r+T_{r+1}+T_{r+2} & ={ }^7 C_1+{ }^7 C_2+{ }^7 C_3 \\\\ & =7+21+35=63 \end{aligned} $$