Question

Calculate \(\sum_{i=0}^{19} 3 \cdot(-2)^{i-1}\)

Answer 1

\(\begin{aligned} \sum_{i=0}^{19} 3 \cdot(-2)^{i-1} & =\left[3 \cdot(-2)^{0-1}\right]+\left[3 \cdot(-2)^{1-1}\right]+\left[3 \cdot(-2)^{2-1}\right]+\ldots \ldots+\left[3 \cdot(-2)^{19-1}\right] \\ & =-\frac{3}{2}+3+(-6)+\ldots .+786432\end{aligned}\)

This is a geometric series with \(a=-\frac{3}{2}\) and \(r=-2\) The number of the terms is:
\[
\begin{aligned}
& n=19-0+1=20 \quad \text { ( Number of terms = Top }-\text { Bottom }+1) \\
& \sum_{i=0}^{19} 3 \cdot(-2)^{i-1} \Rightarrow \mathrm{S}_{20}=?
\end{aligned}
\]
\(\mathrm{S}_n=\frac{a\left(r^n-1\right)}{r-1} \quad\) (state the formula and substitute \(a=-\frac{3}{2}, r=-2\) and \(n=20\) )
\[
\therefore S_{20}=\frac{\left(-\frac{3}{2}\right)\left((-2)^{20}-1\right)}{(-2)-1}=524287,5
\]