×

# Inductive Proof of Exponential Sum

We will inductively prove: $$\sum_{i=0}^{n-1}ia^i = \frac{a-na^n+(n-1)a^{n+1}}{(1-a)^2}$$ where $$a \in \mathbb{R}$$, $$a \neq 1$$.

We establish a base case as $$n=1$$:

$$n=1$$ $$\Rightarrow$$ $$\sum_{i=0}^{0}ia^i = \frac{a-a+0}{(1-a)^2}$$

$$\Rightarrow\ 0=0$$

The base case clearly holds. Now we apply the inductive case $$n=k$$:

$$n=k$$ $$\Rightarrow$$ $$\sum_{i=0}^{k-1}ia^i = \frac{a-ka^k+(k-1)a^{k+1}}{(1-a)^2}$$

$$\Rightarrow$$ $$ka^k +\sum_{i=0}^{k-1}ia^i = ka^k +\frac{a-ka^k+(k-1)a^{k+1}}{(1-a)^2}$$ (By hypothesis)

$$\Rightarrow$$ $$\sum_{i=0}^{k}ia^i = \frac{ka^k(1-a)^2 + a-ka^k+(k-1)a^{k+1}}{(1-a)^2}$$

$$\Rightarrow$$ $$\sum_{i=0}^{k}ia^i = \frac{-ka^{k+1} - a^{k+1} +a +ka^{k+2}}{(1-a)^2}$$

$$\Rightarrow$$ $$\sum_{i=0}^{k}ia^i= \frac{a-(k+1)a^{k+1}+ka^{k+2}}{(1-a)^2}$$

So we can conclude that if the statement is true for some $$n=k$$, then it is true for $$n=k+1$$. The proof follows by induction.

QED

Note by Ethan Robinett
3 years, 4 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Do you know another way to get the summation directly?

Hint: Differentiate.

Staff - 3 years, 4 months ago

Let's see if I can get this right. Starting with $$\displaystyle \sum_{i=0}^{n-1}a^i= \frac{1-a^n}{1-a}$$ and taking your hint:

$$\frac{d}{da}\displaystyle \sum_{i=0}^{n-1}a^i = \frac{d}{da} \frac{1-a^n}{1-a}$$

$$\Rightarrow$$ $$\displaystyle \sum_{i=1}^{n-1}ia^{i-1} = \frac{-na^{n-1}}{1-a} +\frac{1-a^n}{(1-a)^2}$$

$$\Rightarrow$$ $$\frac{1}{a}\displaystyle \sum_{i=1}^{n-1}ia^{i} = \frac{-na^{n-1}}{1-a} +\frac{1-a^n}{(1-a)^2}$$

$$\Rightarrow$$ $$\displaystyle \sum_{i=1}^{n-1}ia^{i} = \frac{-na^n +na^{n+1} +a -a^{n+1}}{(1-a)^2}$$

Since the term corresponding to $$i=0$$ on the left is just zero, we may write:

$$\displaystyle \sum_{i=0}^{n-1}ia^{i} = \frac{-na^n +na^{n+1} +a -a^{n+1}}{(1-a)^2}$$

Which is equivalent to the above. I left out a few of the algebraic details but I think you get the point.

- 3 years, 4 months ago