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

$$\begin{gathered} { }^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}-1}{:}^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}}{:}^{\mathrm{n}+2}{\mathrm{C}}_{\mathrm{r}+1}=1:3:5 \\ \frac{1}{(\mathrm{n}+3-\mathrm{r})!(\mathrm{r}-1)!}:\frac{1}{(\mathrm{n}+2-\mathrm{r})!\mathrm{r}!}:\frac{1}{(\mathrm{n}+1-\mathrm{r})!(\mathrm{r}+1)!}=1:3:5 \\ \frac{1}{(\mathrm{n}+3-\mathrm{r})(\mathrm{n}+2-\mathrm{r})}:\frac{1}{(\mathrm{n}+2-\mathrm{r})\mathrm{r}}:\frac{1}{(\mathrm{r}+1)\mathrm{r}}=1:3:5 \\ \frac{\mathrm{r}}{\mathrm{n}+3-\mathrm{r}}=\frac{1}{3},\frac{\mathrm{r}+1}{\mathrm{n}+2-\mathrm{r}}=\frac{3}{5} \\ \mathrm{n}+3=4\mathrm{r},3\mathrm{n}+1=8\mathrm{r} \\ \therefore \mathrm{n}=5,\mathrm{r}=2 \\ \mathrm{Sum} {=}^7{\mathrm{C}}_1{+}^7{\mathrm{C}}_2{+}^7{\mathrm{C}}_3=7+21+35=63 \end{gathered}$$