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Is there a way to sum up consecutive powers like this when we know the value of $a$ using a closed form expression or do we actually have to add up each one?

By closed form I mean in the same way that the sum of numbers $1$ to $n$ is $n(n+1)/2$.

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2 Answers 2

up vote 3 down vote accepted

It's called a geometric sum.

$$a + a^2 + \cdots + a^n = \frac{a - a^{n+1}}{1 - a}.$$

If it's infinite, then we call it a geometric series and we have

$$a + a^2 + \cdots = \frac{a}{1 - a}$$

though this second one only converges for $|a| < 1$. The proof is similar to the formula you mention. Let $S = a + a^2 + \cdots + a^n$. Now, subtract $S - aS$.

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let $$S_n=a+a^2+...+a^n$$ then $$aS_n=a^2+a^3+...+a^n+a^{n+1}$$ from first equation subtract the second then $$S_n-aS_n=a-a^{n+1}$$ $$S_n(1-a)=a-a^{n+1}$$ $$S_n=\frac{a-a^{n+1}}{1-a}$$

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