# Solving recurrence relation $4a_n=\sum\limits_{k=1}^{n-1} a_k$

I'm trying to solve the recurrence equation $$\begin{cases}4a_n=\displaystyle\sum_{k=1}^{n-1}a_k\\[1ex]a_1=1\end{cases}$$ What I considered doing was subtracting $4a_{n-1}$ from $4a_n$: \begin{align*} 4a_n&=a_1+a_2+\cdots+a_{n-2}+a_{n-1}\\ 4a_{n-1}&=a_1+a_2+\cdots+a_{n-2}\\ \hline 4(a_n-a_{n-1})&=a_{n-1} \end{align*} This reduces to solving $$a_n=\frac{5}{4}a_{n-1}$$ which suggests my solution should be $a_n=\left(\dfrac{5}{4}\right)^{n-1}$. However, this doesn't agree with at least one value of $a_n$ for $n>1$: $$4a_2=\sum_{k=1}^1a_k=1\implies a_2=\frac{1}{4}\neq\frac{5}{4}$$ I'm not so interested in finding the actual correct solution as I am in figuring out why this particular approach doesn't seem to work.

• Your derivation of $a_n = \frac{5}{4} a_{n-1}$ uses $a_{n-1} = a_1 + \cdots + a_{n-2}$ which is valid for $n-1 \ge 2$. So your conclusion $a_n = \frac{5}{4} a_{n-1}$ only works when $n \ge 3$. – achille hui Apr 18 '16 at 20:45
• @achillehui I see, thanks for the prompt reply. It seems shifting the index, as in Jack's answer, is the (only?) way to account for $n\ge2$. – user170231 Apr 18 '16 at 21:53

Since $4a_n = \sum_{k=1}^{n-1}a_k$ it follows that $4a_{n+1}-4a_n = a_n$, so $a_{n+1}=\frac{5}{4}a_n$ for any $n\geq 2$.
It follows that $a_1=1$ and $a_n = \frac{5^{n-2}}{4^{n-1}}$ for any $n\geq 2$.