I’ve got to prove that
$\sum_{n=1}^{15} 3(\frac 23)^n -1 = -9.014$ to 4sf
But keep on getting an answer that is out by about 3 (-6.001) Please could someone help and show me the right steps. I’ve tried separating it out into an arithmetic and geometric but I don’t think that I’ve done it right.
\begin{align} & \sum_{n=1}^{15} \biggl[3\biggl(\frac 23 \biggl)^n -1 \biggl] \\ = & 3\sum_{n=1}^{15} \biggl(\frac 23 \biggl)^n -\sum_{n=1}^{15}1 \\ = & 3 \times \frac{\frac 23 - \bigl(\frac 23 \bigl)^{16}}{1-\frac 23} - 15 \\ = & 9 \biggl( \frac 23 - \Bigl(\frac 23 \Bigl)^{16} \biggl) - 15\\ = & -9.01370... \end{align}
In general, the way to sum $a+a^2+a^3+ \cdots + a^k$ is to let
$$S=a+a^2+a^3+ \cdots + a^k \;$$
Then
$$aS=a^2+a^3+a^4+ \cdots + a^{k+1} \;$$
Subtracting the two equations, we get
$$S-aS = a-a^{k+1}$$
where $S-aS = S(1-a)$, so
$$S = \frac{a-a^{k+1}}{1-a}$$
In this case, $a=\frac 23$ and $k = 15$.