Back

This section is aimed at students in upper secondary education in the Danish school system, some objects will be simplified some details will be omitted.

Annuities

An annuity is a term-wise deposit of money into a capital, which then can accrue or reduce interest depending on whether it's savings or a loan. I will start with the most general case where we have an unknown initial capital and annuity $C_0,A\in\mathbb{R}$ of which the first can be both negative and positive representing loans and withdrawals and savings and payments respectively.

The following result on the partial sum of a geometric series, which technically holds only for $x\neq1$ will be useful for the derivation of the annuity formula.

Geometric Series Lemma

$$\sum_{i=0}^nx^i=\frac{x^{n+1}-1}{x-1}$$

Proof

Consider \begin{align} &&S=&\sum_{i=0}^nx^i=1+x+x^2+\cdots+x^n\\ \implies&&xS=&x+x^2+\cdots+x^n+x^{n+1}\\ &&=&-1+1+x+x^2+\cdots+x^n+x^{n+1}\\ &&=&-1+S+x^{n+1}\\ \implies&& xS-S=&x^{n+1}-1\\ \implies&& S=&\frac{x^{n+1}-1}{x-1} \end{align}

The Annuity Formula Theorem

The capital after $n\in\mathbb{N}$ terms with a term-wise interest $r\in[0,1]$ is $$C_n=\left(C_0+\frac{A}{r}\right)(1+r)^n-\frac{A}{r}$$

Proof

At the start of the process, we just have the initial capital $C_0$. After one term we will have interest on that capital and one annuity, $C_1=C_0(1+r)+A$, after the second term we will have $$C_2=(C_0(1+r)+A)(1+r)+A=C_0(1+r)^2+A(1+r)+A$$ I will now show that by induction we have: $$C_n=C_0(1+r)^n+A(1+r)^{n-1}+A(1+r)^{n-2}+\cdots+A(1+r)+A$$ First off, assume that it works for $C_k$ and lets investigate $C_{k+1}$. \begin{align} C_{k+1}=&C_k\cdot(1+r)+A\\ =&(C_0(1+r)^k+A(1+r)^{k-1}+A(1+r)^{k-2}+\cdots+A(1+r)+A)(1+r)+A\\ =&C_0(1+r)^{k+1}+A(1+r)^k+A(1+r)^{k-1}+\cdots+A(1+r)+A\\ \end{align} Therefore, we have: \begin{align} C_n=&C_0(1+r)^n+A(1+r)^{n-1}+A(1+r)^{n-2}+\cdots+A(1+r)+A\\ =&C_0(1+r)^n+A(1+(1+r)+(1+r)^2+\cdots+(1+r)^{n-1})\\ =&C_0(1+r)^n+A\frac{(1+r)^{n-1+1}-1}{(1+r)-1}\\ =&C_0(1+r)^n+A\frac{(1+r)^n-1}{r}\\ =&C_0(1+r)^n+\frac{A}{r}((1+r)^n-1)\\ =&\left(C_0+\frac{A}{r}\right)(1+r)^n-\frac{A}{r} \end{align}

∎

Savings Corollary

You have annuity savings when $A,C_0\geq0$ but lets assume that $C_0=0$ then we have the following formula $$C_n=A\frac{(1+r)^n-1}{r}$$

Loan Corollary

You have an annuity loan when $C_0< 0$ and $A\geq0$. Lets assume that your goal is to pay back the loan, then $C_n=0$ and you can write the formula as $$A=C_0\frac{r}{1-(1+r)^{-n}}$$