# Proof of Sum of Geometric Series by Mathematical Induction

## Considerations of Sum of Geometric Series

The sum of geometric series is defined using r, the common ratio and n, the number of terms. The common could be any real numbers with some exceptions; the common ratio is 1 and 0.

If the common ratio is 1, the series becomes the sum of constant numbers, so the series cannot be exactly referred to as a geometric series. For example, if the first term is 5, and the common ratio is 1, then the series becomes $5 + 5 + 5 + 5 + \cdots + 5$, so the sum of this series would be the multiply of 5 and the number of terms. It does not need to use any specific formula to evaluate the sum.

If the common ratio is zero, then the series becomes $5 + 0 + 0 + \cdots + 0$, so the sum of this series is simply 5.
Thus our assumptions of finding the sum of geometric series are for any real number, where $r\ne 1$ and $r \ne 0$, where $r =$ the common ratio.

## Sum of Geometric Series Formula

$$a+ar+ar^2+ar^3 + \cdots + ar^n = \displaystyle \frac{a(r^{n+1}-1)}{r-1} \text{ or } \frac{a(1-r^{n+1})}{1-r}$$

## Patterns of Geometric Series

### Sum of the first two terms

\begin{align} 1 – r^{2} &= (1+r)(1-r) \\ \displaystyle \therefore 1+r &= \frac{1 – r^2}{1-r} \cdots (1) \end{align}

### Sum of the first three terms

\begin{align} 1-r^3 &= (1+r+r^2)(1-r) \\ \therefore 1+r+r^2 &= \displaystyle \frac{1-r^3}{1-r} \cdots (2) \end{align}

### Sum of the first four terms

\begin{align} 1-r^4 &= (1+r^2)(1-r^2) \\ &= (1+r^2)(1+r)(1-r) \\ &= (1+r + r^2 + r^3)(1-r) \\ \therefore 1+r + r^2 + r^3 &= \displaystyle \frac{1-r^4}{1-r} \cdots (3) \end{align}

### Sum of the first five terms

\begin{align} 1+r + r^2 + r^3 &= \displaystyle \frac{1-r^4}{1-r} \cdots (3) \\ (1+r + r^2 + r^3)(1-r) &= 1-r^4 \\ (1+r + r^2 + r^3) – (1+r + r^2 + r^3)r &= 1-r^4 \\ (1+r + r^2 + r^3) – \displaystyle \frac{1-r^4}{1-r} \times r &= 1-r^4 \\ (1+r + r^2 + r^3) – \frac{r-r^5}{1-r} &= 1-r^4 \\ r + r^2 + r^3 + r^4 &= \frac{r-r^5}{1-r} \\ 1 + r + r^2 + r^3 + r^4 &= 1 + \frac{r-r^5}{1-r} \\ &= \frac{1-r + r – r^5}{1-r} \\ \therefore 1 + r + r^2 + r^3 + r^4 &= \frac{1- r^5}{1-r} \cdots (4) \end{align}

### Sum of the first $n$ terms

\begin{align} 1 + r + r^2 + r^3 + \cdots + r^{n-1} &= \frac{1- r^n}{1-r} \end{align}

### Sum of the first $n+1$ terms

\begin{align} 1 + r + r^2 + r^3 + \cdots + r^{n-1} + r^n &= \frac{1- r^{n+1}}{1-r} \end{align}

## Proof of Sum of Geometric Series by Mathematical Induction

Now, we will be proving the sum of geometric series formula by mathematical induction.

$\displaystyle 1 + r + r^2 + r^3 + \cdots + r^n = \frac{1-r^{n+1}}{1-r}$

### Step 1

Show it is true for $n=1$.
\begin{align} \text{LHS} &= 1+r \\ \text{RHS} &= \displaystyle \frac{1-r^2}{1-r} \\ &= \frac{(1+r)(1-r)}{1-r} \\ &= 1+r \\ \text{LHS} &= \text{RHS} \end{align}
Therefore the formula is true for $n = 1$.

### Step 2

Assume the formula is true for $n=k$.
That is, $\displaystyle 1 + r + r^2 + r^3 + \cdots + r^k = \frac{1-r^{k+1}}{1-r}$ .

### Step 3

Show the formula is true for $n=k+1$.
That is, $\displaystyle 1 + r + r^2 + r^3 + \cdots + r^k + r^{k+1} = \frac{1-r^{k+2}}{1-r}$ .

\require{AMSsymbols} \displaystyle \begin{align} \text{LHS} &= \bbox[yellow]{1 + r + r^2 + r^3 + \cdots + r^k} + r^{k+1} \\ &= \bbox[yellow]{\frac{1-r^{k+1}}{1-r}} + r^{k+1} &\text{by the assumption} \\ &= \frac{1 – r^{k+1} + r^{k+1} – r^{k+2}}{1-r} &\text{single fraction} \\ &= \frac{1 – r^{k+2}}{1-r} \\ &= \text{RHS} \end{align}

Therefore the formula is true for $n= k+1$.
Hence, the formula is true for all positive integers $n \ne 1$.

The sum of the geometric series is obtained by $$a+ar+ar^2+ar^3 + \cdots + ar^n = \displaystyle \frac{a(r^{n+1}-1)}{r-1} \text{ or } \frac{a(1-r^{n+1})}{1-r}$$, where $n$ is the number of terms, $a$ is the first term, and $r$ is the common ratio of the geometric series
Geometric series is a number sequence connected by adding subsequent terms generated by multiplying common ratio, such as $1 + 2 + 4 + 8 + \cdots + 256$.