Objective: To find present value of a 10 year annuity immediate with level payments of 2 for the first year, 4 for the second year and 6 for the third year. Given annual effective interest rate of 5%.
My idea is to first consider the increasing amount as 24 dollars each year and compute the monthly interest rate as 0.05/12 = 0.00416667. Plugging all this information into an increasing annuity formula is getting me a final answer of 959.79 which is wrong as the expected answer should be $966.44.
equation i am using is : 24*[(((1.0511618979^-10)-1)/0.04867176)-10(1.0511618979)^-10]/0.05
Addendum added to further refine the formula for the present value of a rising annuity.
I was able to reverse-engineer a computation of $966.4356075$ as follows:
Assume that the interest rate is $i$, where $(1 + i)^{12} = 1.05.$ Note that this formula implies that $i \approx 0.004074124$.
Let $\displaystyle v = \frac{1}{1+i}.$
Then, the present value of a payment of $(1)$, made $k$ months in the future is $v^k$.
Assume that the payments are $2$ for each month for the first $12$ months, then $4$ for each month for the next $12$ months, and so on. This implies that the last $12$ payments will be $20$ each.
The problem permits some shortcuts.
For $r,s \in \Bbb{Z^+} ~: r < s$, you have that
$\displaystyle \sum_{k=r}^s v^k = v^r\left(1 + v + v^2 + \cdots + v^{s-r}\right) = v^r \times \frac{1 - v^{(s+1-r)}}{1-v}.$
The present value of the payments equals
$\displaystyle \left(2 \times \sum_{k=1}^{12} v^k\right) + \left(4 \times \sum_{k=13}^{24} v^k\right) + \left(6 \times \sum_{k=25}^{36} v^k\right)$
$\displaystyle + \left(8 \times \sum_{k=37}^{48} v^k\right) + \left(10 \times \sum_{k=49}^{60} v^k\right) + \left(12 \times \sum_{k=61}^{72} v^k\right)$
$\displaystyle + \left(14 \times \sum_{k=73}^{84} v^k\right) + \left(16 \times \sum_{k=85}^{96} v^k\right) + \left(18 \times \sum_{k=97}^{108} v^k\right)$
$\displaystyle + \left(20 \times \sum_{k=109}^{120} v^k\right).$
Let $u = v^{12}$.
Then, the present value can be more easily expressed as
$$2 \times \frac{1-u}{1-v} \times [v] \times (1 + 2u + 3u^2 + 4u^3 + \cdots + 10u^9).\tag1$$
This formula led to the computation shown in the first line of my answer.
Addendum
The formula in (1) above includes the factor of
$(1 + 2u + 3u^2 + \cdots + 9u^8 + 10u^9).$
An expression of this form is often encountered when computing the present value of a rising annuity. It therefore seems reasonable to further refine the formula in (1) above by deriving a shortcut for the above factor.
For $u \in \Bbb{R}, ~u \neq 1, ~n \in \Bbb{Z_{\geq 0}}:$
let $S(n,u)$ denote $\displaystyle \left[1 + u + u^2 + \cdots + u^n\right] = \frac{1-u^{n+1}}{1 - u}~~~~$ and
let $R(n,u)$ denote $\displaystyle \left[1 + 2u + 3u^2 + \cdots + nu^{n-1} + (n+1)u^n\right].$
The goal in this Addendum will be to derive a closed form expression for $R(n,u)$ and then apply it to the formula in (1) above.
$R(n,u)$ may be re-expressed as:
$\left(1 + u + u^2 + \cdots + u^n\right)$
$+~ u\left(1 + u + u^2 + \cdots + u^{n-1}\right)$
$+~ u^2\left(1 + u + u^2 + \cdots + u^{n-2}\right)$
$+~ \cdots $
$+~ u^{n-1}\left(1 + u\right)$
$+~ u^n\left(1\right).$
Using the formula for $S(n,u)$ above, $R(n,u)$ equals:
$\displaystyle \left(\frac{1}{1-u}\right) \times \left[\left(1 - u^{n+1}\right)\right]$
$\displaystyle +~ \left(\frac{1}{1-u}\right) \times \left[u\left(1 - u^{n}\right)\right]$
$\displaystyle +~ \left(\frac{1}{1-u}\right) \times \left[u^2\left(1 - u^{n-1}\right)\right]$
$+~ \cdots$
$\displaystyle +~ \left(\frac{1}{1-u}\right) \times \left[u^{n-1}\left(1 - u^{2}\right)\right]$
$\displaystyle +~ \left(\frac{1}{1-u}\right) \times \left[u^{n}\left(1 - u\right)\right].$
This equals:
$\displaystyle \left(\frac{1}{1-u}\right) \times \left[\left(1 + u + u^2 + \cdots + u^{n-1} + u^n\right) - (n+1)u^{n+1}\right]$
$\displaystyle =~ \left(\frac{1}{1-u}\right) \times \left[\left(\frac{1 - u^{n+1}}{1-u}\right) - (n+1)u^{n+1}\right].$
Therefore,
$$R(n,u) ~=~ \frac{1 - u^{n+1}}{(1-u)^2} ~-~ \frac{(n+1)u^{n+1}}{1-u}.\tag2 $$
In (1) above, the factor of $R(n,u)$ appears with $n = 9$. Therefore, the formula in (1) above may be re-expressed as:
$\displaystyle (1 + i) = (1.05)^{(1/12)}.$
$\displaystyle v = \frac{1}{1 + i}$.
$\displaystyle u = v^{12}.$
The present value of the rising annuity equals:
$$2 \times \frac{1-u}{1-v} \times [v] \times \left[\frac{1 - u^{10}}{(1-u)^2} ~-~ \frac{(10)u^{10}}{1-u}\right].\tag3$$