- TODO
- Add product of a range of geometric progression.
Geometric progression is sequence of numbers where the ratio between any two consecutive numbers is constant.
For example, is a geometric progression with constant .
Sum of a range of geometric progression
\dfrac{bk - a}{k - 1}, &k \neq 1,\\ an, &k = 1 \end{cases}$$ Where: - $a$: The first number - $b$: The last number - $k$: Ratio between the consecutive numbers ### Proof Let $a, k \in \mathbb R, n \in \mathbb N, n > 0$. Define $S_n$ byS_n = \sum_{i = 0}^n ak^i = a + ak + ak^2 + \dots + ak^{n-1} + ak
Consider all cases for $k$: - $k \neq 1$ Multiplying $S_n$ by $k$, we get $$ kS_n = \sum_{i = 0}^n ak^{i + 1} = ak + ak^2 + ak^3 + \dots + ak^{n} + ak^{n+1} $$ Solving $k S_n - S$, we get $$\begin{align}k S_n - S_n &= (ak + ak^2 + ak^3 + \dots + ak^{n} + \textcolor{yellow}{ak^{n+1}}) - (\textcolor{yellow}a + ak + ak^2 + \dots + ak^{n-1} + ak^n)\\ &= ak - ak + ak^2 - ak^2 + \dots + ak^n - ak^n + \textcolor{yellow}{ak^{n+1} - a}\\ &= \textcolor{yellow}{ak^{n+1} - a} \end{align}$$ Define $b$ by $b = ak^{n}$, then $ak^{n+1} = ak^{n}\cdot k = b\cdot k$, thus $$\begin{array}{rrl} &kS_n - S_n &= bk - a\\ \iff &S_n(k - 1) &= bk - a\\ \iff &S_n &= \frac{bk - a}{k - 1} \end{array}$$ - $k = 1$ Notice that $S_n = \sum_{i = 0}^n a\cdot 1 = an$ From all cases of $k$, we get $$S_n = \begin{cases} \dfrac{bk - a}{k - 1}, &k \neq 1,\\ an, &k = 1 \end{cases}$$