I was given the following sum to solve:
$E[t_n] = \sum_{t = 1}^{n+1} t \cdot Pr(t_n = t)$
$= (n+1) \cdot (0.99)^n + 0.01 \cdot \sum_{t = 1}^{n} t \cdot (0.99)^{t - 1} $
As well as the following solution to this sum:
$ S = \sum_{t = 1}^{n}t \cdot 0.99^{t-1} = 1 + 2\cdot0.99+3\cdot0.99^2 + ... + n \cdot 0.99^{n - 1} $
$ 0.99S = \sum_{t = 1}^{n}t \cdot 0.99^t = 1 \cdot 0.99 + 2\cdot0.99^2 +3\cdot0.99^3 + ... + (n-1) \cdot 0.99^{n - 1} + n \cdot 0.99^n $
take the difference of the above two equations
$ 0.01S = \sum_{i = 0}^{n-1} 0.99^i - n \cdot 0.99^n = \frac{1-0.99^n }{1-0.99} - n \cdot 0.99^n $
$ = 100 - (100 + n) \cdot 0.99^n $
where $\frac{1-0.99^n }{1-0.99}$ is the sum of a geometric series.
However, I do not understand what formula was used to solve the sum as well as how the following line was arrived at or what $i$ is:
$ 0.01S = \sum_{i = 0}^{n-1} 0.99^i - n \cdot 0.99^n = \frac{1-0.99^n }{1-0.99} - n \cdot 0.99^n $
We have $$S = 1 + 2\cdot 0.99 + 3 \cdot 0.99^2 + \dots + n \cdot 0.99^{n-1} $$ Multiplying by $0.99$ gives $$0.99S = 1\cdot 0.99 + 2 \cdot 0.99^2 + \dots + n \cdot 0.99^{n} $$ Now we're going to subtract the second equation from the first, combining the terms with the same power of $0.99$. This gives $$ \begin{align} S - 0.99S &= 1 + (2-1)\cdot 0.99 + (3-2)\cdot 0.99^2 + \dots + (n-(n-1))\cdot0.99^{n-1} - n\cdot 0.99^n \\ 0.01S &= 1 + 1\cdot0.99 + 1\cdot0.99^2 + \dots + 1\cdot0.99^{n-1} - n\cdot 0.99^n \end{align} $$ $i$ is just a placeholder variable to write all the terms on the right-hand side (except the last) in a single sum. These terms are a geometric series, and the solution you posted uses the formula for a sum of a geometric series and proceeds from there.