$$a_n=n+n \cdot 5^n \quad n \geq 0$$ $$b_n=\sum_{k=0}^{n-1} a_k \quad n \geq 1, b_0=0$$ Find explicit expression for: $$\sum_{n \geq 0} b_n x^n$$
So we have $\sum_{n \geq 0} \Big( \sum_{k=0}^{n-1} k + k \cdot 5^k \Big) x^n$.
Should somehow I use Cauchy product here? I've calculated just the inner sum of $k$, so far: $\sum_{k=0}^{n-1} k=\frac{1}{2} (n^2-n)$.
$$b_n=\sum_{k=0}^{n-1} a_k=\sum_{k=0}^{n-1} (k+k \cdot 5^k)=\sum_{k=0}^{n-1}k +5\sum_{k=0}^{n-1}k\cdot 5^{k-1}$$ The first sum is simple $$\sum_{k=0}^{n-1}k=\frac{1}{2} (n-1) n$$ For the second one, consider $$S_n=\sum_{k=0}^{n-1}k\cdot x^{k-1}=\frac d {dx}\Big(\sum_{k=0}^{n-1} x^k\Big)=\frac d {dx}\Big(\frac{x^n-1}{x-1} \Big)$$ When you finish, replace $x$ by $5$; this will give the expression of $b_n$.
Then, you need to find the corresponding generating function corresponding to $\sum_{n=0}^\infty b_n x^n$
