I am now learning math to be a actuary in korea.
I don't understand how this expansion can happen in this equation. I've looked on my calculus book and I couldn't find anything useful.
https://i.stack.imgur.com/M6Ee9.jpg
https://i.stack.imgur.com/mSC0H.jpg
and v = 1/(1+i)
How this expansion can happen? In my book, the numbers are keep going down like n*(n-1)(n-2)... in expansion but in this equation numbers are going up in multiplication
$\require{enclose}$ $v = 1/(1+i)$ is the effective periodic present value discount factor. In other words, if $i$ is the effective periodic (usually annual) interest rate, then $v$ represents the present value of $1$ paid at time $t = 1$.
For instance, if you borrow $100$ today at an effective annual interest rate of $i = 0.05$, and must pay it back one year from now, the amount that you must pay back is $100(1+i) = 105$. The future payment is discounted by $v$ to get the present value of that payment: $$105v = 105 \cdot \frac{1}{1 + 0.05} = 100.$$
If you have an annuity that pays $1$ at the end of each year for $n$ years, then the present value of the first payment is $v$. The present value of the second payment is $v^2$, because it takes $2$ years to receive it. So the total present value of the payment stream is just $$PV = v + v^2 + v^3 + \cdots + v^n.$$ When $i > 0$, the present value of each payment decreases because although you receive the same nominal value $1$, the time value of money means that the longer you have to wait to receive payment, the less value it has to you right now. If you have a choice between being paid $100$ tomorrow, or paid $100$ in five years, which would you choose?
The expression $a_{\enclose{actuarial}{n} i}$ is actuarial notation for the present value of an annuity-immediate of $1$ per year for $n$ years at an effective annual interest rate $i$. This is what we had written above, so $$a_{\enclose{actuarial}{n} i} = v + v^2 + \cdots + v^n = \frac{1 - v^n}{i},$$ where the right-hand side of the equation is just an application of the formula for a finite geometric series with common ratio $v$.
Where do the other formulas come from? We can also write
$$\begin{align} a_{\enclose{actuarial}{n} i} &= \frac{1 - v^n}{i} \\ &= \frac{1 - (1+i)^{-n}}{i} \\ &= \frac{1}{i} \left(1 - \sum_{k=0}^\infty \binom{-n}{k} i^k \right) \tag{$*$} \\ &= \frac{1}{i} \left(1 - 1 - (-n)i - \frac{(-n)(-n-1)}{2!} i^2 - \frac{(-n)(-n-1)(-n-2)}{3!} i^3 - \cdots \right) \\ &= \frac{1}{i} \left( ni - \frac{n(n+1)}{2!} i^2 + \frac{n(n+1)(n+2)}{3!} i^3 - \cdots \right) \\ &= n - \frac{n(n+1)}{2!} i + \frac{n(n+1)(n+2)}{3!} i^2 - \cdots. \end{align}$$ Here, we used the binomial series formula at the step $(*)$. This formula is not commonly used.
Then the last formula, to my knowledge, is obtained by inverting the previous series expansion. In other words, the coefficients are found by solving for
$$\frac{1}{n}(c_0 + c_1 i + c_2 i^2 + \cdots )\left(n - \frac{n(n+1)}{2} i + \frac{n(n+1)(n+2)}{6} i^2 + \cdots \right) = 1.$$ I don't think there is an elementary closed form formula for the coefficients $c_k$. For example,
$$\begin{align} c_3 &= \frac{1-n^2}{24} \\ c_4 &= \frac{-19+20n^2-n^4}{720} \\ c_5 &= \frac{9-10n^2+n^4}{480} \\ c_6 &= \frac{-863+1008n^2-147n^4+2n^6}{60480},\\ &\vdots \end{align}$$
I honestly don't see any meaningful use of this formula.