Increasing and decreasing annuity Going in circles.....
Find an expression for the present value of an annuity-immediate where payments start at 1, increase by 1 each period up to a payment of $n$, and then decrease by 1 each period up to a final payment of 1.
Okay... We know that for the first $n$ payments, we have an increasing annuity $(Ia)_n$ which equals $\frac{\ddot{a}_n-n\nu^n}{i}$ and we have a decreasing annuity $(Da)_{n-1}$ which equals $\frac{(n-1)-a_{n-1}}{i}$, but because this annuity starts decreasing at period n to get the present value we have to multiply $(Da)_{n-1}$ by $\nu^n$.
Thus, our compound annuity is equal to:
$\frac{\ddot{a}_n-n\nu^n}{i}$+$\nu^n\frac{(n-1)-a_{n-1}}{i}$ = $\frac{(1+i)a_n-n\nu^n}{i}$+$\frac{\nu^n(n-1)-\nu^na_{n-1}}{i}$ = $\frac{(1+i)a_n-\nu^n-\nu^na_{n-1}}{i}$ = $\frac{(1+i)a_n-\nu^n(1+a_{n-1})}{i}$
and now I'm stuck...  Finan gives the answer as $a_n\cdot{\ddot{a}_n}$

The present values of the payments can be written as 
$$\begin{align}
 PV &= a_n + a_nv + a_n v^2 + a_n v^3 +a_n v^4 + \ldots +a_n v^n\\
 &  = a_n(1+v+v^2+v^3+v^4+\ldots + v^n)\\
& = a_n \cdot{\ddot{a}_n}
\end{align}$$ 
Alternatively, you could write the present value as $$
Ia_n + Da_{n-1} v^n$$ and exploit the relationships between $a_n$ and $\ddot{a}_n$ . i.e .
$$\begin{align}
Ia_n + Da_{n-1} v^n & = \frac{\ddot{a}_n - nv^n}{i} + \frac{n-1-a_{n-1}}{i}v^n\\
& = \frac{\ddot{a}_n-nv^n+nv^n-v^n-a_{n-1}v^n}{i}\\
& = \frac{\ddot{a}-v^n-a_{n-1}v^n}{i}\\
& = \frac{1+a_{n-1}-v^n-a_{n-1}v^n}{i}\\
& = \frac{1-v^n}{i} + \frac{a_{n-1}-a_{n-1}v^n}{i}\\
& = a_n + a_{n-1}\frac{1-v^n}{i}\\
& = a_n + a_{n-1}\cdot a_n\\
& = a_n\left(1+a_{n-1}\right)\\
& = a_n\cdot \ddot{a}_n
\end{align}$$ 
where I have used the fact that $\ddot{a}_n = 1+ a_{n-1}$ and $a_n = \frac{1-v^n}{i}$.
