<s>[INST] Can I get a hint on solving this recurrence relation? I am having trouble solving for a closed form of the following recurrence relation.
$$\begin{align*}
a_n &= \frac{n}{4} -\frac{1}{2}\sum_{k=1}^{n-1}a_k\\
a_1 &= \frac{1}{4}
\end{align*}$$
The first few values are $a_1=\frac{1}{4},a_2=\frac{3}{8},a_3=\frac{7}{16},a_4=\frac{15}{32}...$ So it seems the pattern is $a_n = \frac{2^{n}-1}{2^{n+1}}$, but I have been unable to show this algebraically. Here is what I tried:
$$\begin{align*}
2a_n &= \frac{n}{2} - \sum_{k=1}^{n-1}a_k\\
a_n + \sum_{k=1}^n a_k &= \frac{n}{2}\\
a_{n-1} + \sum_{k=1}^{n-1} a_k &= \frac{n-1}{2} \\
2a_n - a_{n-1} & = \frac{1}{2} \\
a_n = \dfrac{2a_{n-1} + 1}{4}
\end{align*}$$
I am so close, I can taste the closed form. Can someone nudge me in the right direction without giving too much away?
 [/INST] Subtract the formula for $a_{n-1}$ from that for $a_n$
$$
a_n-a_{n-1}=\frac14-\frac12a_{n-1}
$$
Multiply by $2^n$ and bring $a_{n-1}$ from the left to the right
$$
2^na_n=2^{n-2}+2^{n-1}a_{n-1}
$$
Using the formula for the sum of a geometric series, we get
$$
2^na_n=2^{n-1}+C
$$
Plug in $n=1$ to find that $C=-\frac12$. Thus,
$$
a_n=\frac12-\frac1{2^{n+1}}
$$
</s>