4. Let the sequence $\left\{a_{n}\right\}$ be a geometric sequence with the first term $a$ and common ratio $t(t \neq 1)$. Let $b_{n}=1+a_{1}+a_{2}+\cdots+$ $a_{n}, c_{n}=2+b_{1}+b_{2}+\cdots+b_{n}, n \in \mathbf{N}_{+}$. The pairs of real numbers $(a, t)$ that make $\left\{c_{n}\right\}$ a geometric sequence are $\qquad$
4. $(1,2)$.

Since $t \neq 1, b_{n}=1+\frac{a}{1-t}-\frac{a t^{n}}{1-t}$, we have
$$
\begin{array}{l}
c_{n}=2+\left(1+\frac{a}{1-t}\right) n-\frac{a}{1-t}\left(t+t^{2}+\cdots+t^{n}\right) \\
=2+\left(1+\frac{a}{1-t}\right) n-\frac{a\left(t-t^{n+1}\right)}{(1-t)^{2}} . \\
=2-\frac{a t}{(1-t)^{2}}+\left(\frac{1-t+a}{1-t}\right) n+\frac{a t^{n+1}}{(1-t)^{2}} .
\end{array}
$$

Given that $\left\{c_{n}\right\}$ is a geometric sequence, then
$$
\left\{\begin{array}{l}
2-\frac{a t}{(1-t)^{2}}=0, \\
\frac{1-t+a}{1-t}=0 .
\end{array}\right.
$$

Solving these equations, we get $a=1, t=2$.