$$0.01\sum_{x=1}^{30}(0.99)^{x-1} = 1-0.99^{30}$$
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.
$$0.01\sum_{x=1}^{30}(0.99)^{x-1} = 1-0.99^{30}$$
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.
On
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 $$ \displaystyle 0.01\left(\frac{1(1-0.99^{30})}{1-0.99}\right)=0.01\left(\frac{1(1-0.99^{30})}{0.01}\right)=1-0.99^{30}. $$
its geometric series
$$\sum_{n=0}^{d-1}x^n=\frac{1-x^{d}}{1-x}$$
$$0.01\sum_{x=1}^{30}(0.99)^{x-1}=0.01\frac{1-0.99^{30}}{1-0.99}=1-0.99^{30}$$