# Prove by Mathematical Induction? :  sum_(k=0)^n x^k = (1-x^(n+1)) / (1-x)

Aug 6, 2017

We aim to prove by Mathematical Induction that for $n \in \mathbb{N} , n \ge 1$ then:

${\sum}_{k = 0}^{n} {x}^{k} = \frac{1 - {x}^{n + 1}}{1 - x}$

When $n = 1$ the given result gives:

$L H S = {\sum}_{k = 0}^{n} {x}^{k} = {x}^{0} + {x}^{1} = 1 + x$
$R H S = \frac{1 - {x}^{2}}{1 - x} = \frac{\left(1 + x\right) \left(1 - x\right)}{1 - x} = 1 + x$

So the given result is true when $n = 1$

Now, Let us assume that the given result is true when $n = m$, for some $m \in \mathbb{N} , m > 1$, in which case for this particular value of $m$ we have:

${\sum}_{k = 0}^{m} {x}^{k} = \frac{1 - {x}^{m + 1}}{1 - x}$

Adding the next term of the sum gives us:

${\sum}_{k = 0}^{m + 1} {x}^{k} = \left({\sum}_{k = 0}^{m} {x}^{k}\right) + {x}^{m + 1}$

And using the above assumption, we have:

${\sum}_{k = 0}^{m + 1} {x}^{k} = \frac{1 - {x}^{m + 1}}{1 - x} + {x}^{m + 1}$
$\text{ } = \frac{1 - {x}^{m + 1} + \left(1 - x\right) {x}^{m + 1}}{1 - x}$
$\text{ } = \frac{1 - {x}^{m + 1} + {x}^{m + 1} + x {x}^{m + 1}}{1 - x}$
$\text{ } = \frac{1 + x {x}^{m + 1}}{1 - x}$
$\text{ } = \frac{1 + {x}^{m + 2}}{1 - x}$
$\text{ } = \frac{1 + {x}^{\left(m + 1\right) + 1}}{1 - x}$

Which is the given result with $n = m + 1$

So, we have shown that if the given result is true for $n = m$, then it is also true for $n = m + 1$. But we initially showed that the given result was true for $n = 1$ and so it must also be true for $n = 2 , n = 3 , n = 4 , \ldots$ and so on.

Hence, by the process of mathematical induction the given result is true for $n \in \mathbb{N} , n \ge 1$ QED

Supporting Note

The above result should come as no surprise because:

${\sum}_{k = 0}^{n} {x}^{k} = {x}^{0} + {x}^{1} + {x}^{2} + \ldots + {x}^{n}$

Which are the $n + 1$ terms of a GP with:

First Term $a = {x}^{0} = 1$,
Common Ratio $r = x$

And so, using the GP formula, the sum of the first $n$ terms is given by:

${S}_{n} = a \frac{1 - {r}^{n}}{1 - r}$
$\setminus \setminus \setminus \setminus = 1 \frac{1 - {x}^{n + 1}}{1 - x} \setminus \setminus \setminus$ (remember we have $n + 1$ terms)
$\setminus \setminus \setminus \setminus = \frac{1 - {x}^{n + 1}}{1 - x}$ QED