Accumulation of a series of payments A loan is repayable by eight annual payments, starting in one year's time with an interest rate $i$. Payments one to three are half as much as payments four to eight. What is the accumulated value of payments one year before the end of the eight annual payments.
What I thought it would be:
Let $X$ be the payment per year. From year 1 to 3 we pay $X$ Annually, and $2X$ From year 3 to 7, so the accumulated value is $$X + X(1+i) + X(1+i)^2 + 2X(1+i)^3 + 2X(1+i)^4 + 2X(1+i)^5 + 2X(1+i)^6$$, but the correct answer is in fact:
$$2X + 2X(1+i) + 2X(1+i)^2 + 2X(1+i)^3 + X(1+i)^4 + X(1+i)^5 + X(1+i)^6$$
Could someone explain the logic behind this?
 A: You have the correct value of the payments, but it looks like your interest factors are a bit off. The first payment of $X$ takes place in one year. It will sit in the account for six years (since the accumulation takes place one year before the loan is paid off). This gives you that the first payment will accumulate to
\begin{equation*}
X(1+i)^6
\end{equation*}
The next two payments will follow suit, but they will each have a reduced exponent. The accumulated values of those payments are
\begin{align*}
X(1+i)^5\\
X(1+i)^4
\end{align*}
respectively.
The remaining payments are all $2X$, and they grow $3$, $2$, $1$, and $0$ years. This gives us the values
\begin{align*}
2X&(1+i)^3\\
2X&(1+i)^2\\
2X&(1+i)\\
2X&
\end{align*}
Summing the seven values I listed gives you the desired response.
\begin{equation*}
2X+2X(1+i)+2X(1+i)^2+2X(1+i)^3+X(1+i)^4+X(1+i)^5+X(1+i)^6
\end{equation*}
Although, I prefer your order:
\begin{equation*}
X(1+i)^6+X(1+i)^5+X(1+i)^4+2X(1+i)^3+2X(1+i)^2+2X(1+i)+2X
\end{equation*}
