An equality: from one sum to 2 sums.

Question

I have an equality: $$ \ddot{a}_{30} = \frac{1}{0.75}\left( \sum_{k=0}^{\infty} \left(\frac{1}{1.06}\right)^k \left( 1 - \frac{30+k}{120}\right) \right)=\left( \sum_{k=0}^{\infty} \left(\frac{1}{1.06}\right)^k - \frac{1}{90} \sum_{k=0}^{\infty} k\left(\frac{1}{1.06}\right)^k \right) $$

How from $\frac{1}{0.75}\left( \sum_{k=0}^{\infty} \left(\frac{1}{1.06}\right)^k \left( 1 - \frac{30+k}{120}\right) \right)$ we get $\left( \sum_{k=0}^{\infty} \left(\frac{1}{1.06}\right)^k - \frac{1}{90} \sum_{k=0}^{\infty} k\left(\frac{1}{1.06}\right)^k \right) ?$

Because I do not understand where we lost $\frac{1}{0,75}$ in the first sum.

Answer 1

\begin{align}\frac{1}{0.75}\left( \sum_{k=0}^{\infty} \left(\frac{1}{1.06}\right)^k \left( 1 - \frac{30+k}{120}\right) \right)&= \frac{4}{3}\left( \sum_{k=0}^{\infty} \left(\frac{1}{1.06}\right)^k \left( \frac34 - \frac{k}{120}\right) \right) \\&=\left( \sum_{k=0}^{\infty} \left(\frac{1}{1.06}\right)^k \left( 1 - \frac{k}{90}\right) \right) \\&=\left( \sum_{k=0}^{\infty} \left(\frac{1}{1.06}\right)^k - \frac{1}{90} \sum_{k=0}^{\infty} k\left(\frac{1}{1.06}\right)^k \right) \end{align}

The $\frac43$ and the $\frac34$ cancels out.