The question asks to find the sum of the geometric series, given this:
$$\sum_{m=2}^{10} 5^{m-3}$$
I found that the common ratio is $r=5$. The formula that I use to find the sum is $S_n = a\frac{r^n - 1}{r - 1}$.
What do I substitute into n and a?
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Carl Christian
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2Try to write your sum in the form $a(1+r+r^2+\cdots+r^{n-1})$ and then it will become clear. – angryavian Oct 22 '17 at 07:27
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Substitute $a=\frac 15$ and $n=9$.
EDIT : Notice that the first term of the summation is at $n=2$, i.e. the first term is $5^{2-3}=\frac 15.$
Next realize that we are taking summation from $2$ to $10$, so in total we have $(10-2)+1=9$ terms in the summation. (Any list $m,m+1,m+2,...,n$ of integers has $(n-m)+1$ number of terms.)
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You could use $a=\frac1 {5^3} $ and the formula $a\frac {r^{n+1}-5^2}{r-1}$,with $n=10$ and $r=5$, since the series starts with $5^2$...
You get $(\frac 1 {125})\frac {5^{11}-5^2}{5-1}=(\frac 15) \frac {5^9-1}4$... like above. ..