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# 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
2 years, 9 months ago

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Do you know another way to get the summation directly?

Hint: Differentiate. Staff · 2 years, 9 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. · 2 years, 9 months ago