8. Given that $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are both arithmetic sequences, and $a_{11}=32, b_{21}=43$. Let $c_{n}=(-1)^{n} \cdot\left(a_{n}-b_{n}\right)$, the sum of the first 10 terms of the sequence $\left\{c_{n}\right\}$ is 5, and the sum of the first 13 terms is -5, then the sum of the first 100 terms of the sequence $\left\{b_{n}\right\}$ is
Let the common differences of the arithmetic sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$ be $d_{1}, d_{2}$, respectively, then
$$
\begin{array}{l}
c_{2 k-1}+c_{2 k}=-\left(a_{2 k-1}-b_{2 k-1}\right)+\left(a_{2 k}-b_{2 k}\right)=a_{2 k}-a_{2 k-1}+b_{2 k-1}-b_{2 k}=d_{1}-d_{2}, \\
\text { thus }\left\{\begin{array} { l } 
{ 5 ( d _ { 1 } - d _ { 2 } ) = 5 , } \\
{ b _ { 1 3 } - a _ { 1 3 } + 6 ( d _ { 1 } - d _ { 2 } ) = - 5 }
\end{array} \Rightarrow \left\{\begin{array}{l}
b_{1}=a_{1}+1, \\
d_{2}=d_{1}-1 .
\end{array}\right.\right. \\
\text { Substituting into }\left\{\begin{array} { l } 
{ a _ { 1 } + 1 0 d _ { 1 } = 3 2 , } \\
{ b _ { 1 } + 2 0 d _ { 2 } = 4 3 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ a _ { 1 } + 1 0 d _ { 1 } = 3 2 , } \\
{ a _ { 1 } + 2 0 d _ { 1 } = 6 2 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ a _ { 1 } = 2 , } \\
{ d _ { 1 } = 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
b_{1}=3, \\
d_{2}=2 .
\end{array}\right.\right.\right.\right.
\end{array}
$$

Therefore, $b_{n}=2 n+1$. Thus, the sum of the first 100 terms of the sequence $\left\{b_{n}\right\}$ is $101^{2}-1=10200$.