Ratio of three consecutive terms in the binomial expansion of ${\left(1+x\right)}^{n+2}$ is

 ${}^{n+2}C_{r-1}:{}^{n+2}C_r:{}^{n+2}C_{r+1}::1:3:5$

So,

$\Rightarrow \frac{{}^{n+2}C_r}{{}^{n+2}C_{r-1}}=\frac{3}{1}$

$\Rightarrow \frac{n-r+3}{r}=3$

$\Rightarrow n+3=4r ...\left(1\right)$

And,

$\frac{{}^{n+2}C_{r+1}}{{}^{n+2}C_r}=\frac{5}{3}$

$\Rightarrow \frac{n+2-r}{r+1}=\frac{5}{3}$

$\Rightarrow 3n+6-3r=5r+5$

$\Rightarrow 3n+1=8r ...\left(2\right)$

So, $3\left(4r-3\right)+1=8r$

$\Rightarrow 12r-9+1=8r$

$\Rightarrow 4r=8$

$\Rightarrow r=2$

So, $n=5$

Hence, we have

${\left(1+x\right)}^{n+2}={\left(1+x\right)}^7$

Sum of consecutive terms is

${}^{7}C_1+{}^{7}C_2+{}^{7}C_3$

$=7+21+35=63$