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$$0.01\sum_{x=1}^{30}(0.99)^{x-1} = 1-0.99^{30}$$

I wonder if there is a formula for summation and I want to know. Would anyone mind telling me? It would be better for me to solve problems, like the above one.

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its geometric series

$$\sum_{n=0}^{d-1}x^n=\frac{1-x^{d}}{1-x}$$

$$0.01\sum_{x=1}^{30}(0.99)^{x-1}=0.01\frac{1-0.99^{30}}{1-0.99}=1-0.99^{30}$$

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There is no formula for summation as such. Each summation can be written in longhand i.e. the one in your question is equivalent to $$ \displaystyle 0.01(0.99^0+0.99^1+0.99^2+\cdots 0.99^{29}) $$ which is obviously the same as the RHS. Use the sum $S_n$ of the first $n$ terms of a geometric series with $a=1$ and $r=0.99$ to get $$ \displaystyle 0.01\left(\frac{1(1-0.99^{30})}{1-0.99}\right)=0.01\left(\frac{1(1-0.99^{30})}{0.01}\right)=1-0.99^{30}. $$

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