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

## Comments

Sort by:

TopNewestDo you know another way to get the summation directly?

Hint: Differentiate. – Calvin Lin Staff · 2 years, 7 months ago

Log in to reply

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

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

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

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

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 · 2 years, 7 months ago

Log in to reply