Waste less time on Facebook — follow Brilliant.
×

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

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. 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 1

paragraph 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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Do you know another way to get the summation directly?

Hint: Differentiate.

Calvin Lin Staff - 3 years, 4 months ago

Log in to reply

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.

Ethan Robinett - 3 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...