(7) Given the polynomial $(1+x)+(1+x)^{2}+(1+x)^{3}+\cdots+(1+x)^{n}=b_{0}+$ $b_{1} x+b_{2} x^{2}+\cdots+b_{n} x^{n}$, and it satisfies $b_{1}+b_{2}+\cdots+b_{n}=26$, then a possible value of the positive integer $n$ is $\qquad$ .
$(7)$ Taking $x=0$, we get $b_{0}=n$, taking $x=1$, we get
$$
2+2^{2}+2^{3}+\cdots+2^{n}=b_{0}+b_{1}+b_{2}+\cdots+b_{n},
$$

which means
$$
\frac{2\left(1-2^{n}\right)}{1-2}=n+26
$$

or equivalently,
$$
2\left(2^{n}-1\right)=n+26,
$$

then $n=4$ is a possible value.