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I'm working on a question like: find a simple g(n) (big theta) for $ f(n) = \sum _{i=1}^n 5^i $

My working starts with this $\frac{5-5^n}{1-5}$ which is not equivalent to the correct answer from wolfram alpha: $\frac{5}{4}\left(5^n-1\right)$. Step-by-step solutions aren't available for this problem type unfortunately.

(I know that this isn't the final answer for big theta but I'm stuck at this step and need to show how I reach the answer for g(n))

I'd been using the second equation from here on wikipedia.

Charlie Egan
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    Your sum is $5(1+5+\cdots+5^{n-1})$. The sum is 5$\cdot \frac{1-5^n}{1-5}$. Of course we multiply top and bottom by $-1$ to make them both positive. – André Nicolas Jun 01 '14 at 20:11
  • Where can I find the formula for the second sum in your comment? Or, why is the one on Wikipedia not the correct one for me to use here? – Charlie Egan Jun 01 '14 at 20:16
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    The Wikipedia sum is $\sum_0^{n-1} a^i$. Your sum is $\sum_1^n a^i$. But as I observed for your concrete case $a=5$, we have $\sum_1^n a^i=a\sum_0^{n-1}a^i$, and now we can use the Wikipedia formula. – André Nicolas Jun 01 '14 at 20:20
  • Ahh, okay! Wow cheers, that makes a lot more sense now! – Charlie Egan Jun 01 '14 at 20:21

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