$(1+x)^n=1+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\cdots+\binom{n}{n}x^n$
Putting $x=1, 2~$ yields respectively
$2^n=1+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\binom{n}{4}+\cdots+\binom{n}{n} \tag{i}$
and
$3^n=1+\binom{n}{1}\times2+\binom{n}{2}\times2^2+\binom{n}{3}\times2^3+\binom{n}{4}\times2^4+\cdots+\binom{n}{n}\times2^n \tag{ii}$
Adding no$\mathrm{(i)}$ and no$\mathrm{(ii)}$,
$\begin{align} 2^n+3^n&=\scriptsize 2+\binom{n}{1}\times(1+2)+\binom{n}{2}\times(1+2^2)+\binom{n}{3}\times(1+2^3)+\binom{n}{4}\times(1+2^4)+\cdots+\binom{n}{n}\times(1+2^n)\\ &=\scriptsize 2+3\times\binom{n}{1}+5\times\binom{n}{2}+9\times\binom{n}{3}+17\times\binom{n}{4}+\cdots+(2^n+1)\times\binom{n}{n} \end{align}$
Definitely it has $n+1$ terms.
So the correct answer is C.