How do I find the sum of the series?

Question

$$\sum_{k=1}^{7}40 \left( \frac{1}{2}\right)^{k-1} = \frac{635}{8}$$

The image of the orginial eqn is on the link above and so is the answer, but I need help in how to solve it. when I did solve it I got $$\displaystyle 5(\frac{127}{8})$$

What did I do wrong? when solving: $$r=1/2$$ $$Ag=40$$ $$n=7$$ So I plugged in everything into the formula: $$\begin{align} \sum &=Ag \frac{1-r^n}{1-r}\\ &= 40 \frac{1-(1/2)^7}{1-1/2}\\ &=40\frac{1-1/128}{2/2-1/2}\\ &=49\frac{128/128-1/128}{1/2}\\ &=40\frac{127/128}{1/2}\\ &=40\frac{127}{128}\cdot\frac21\\ &=40\frac{127}{64}\cdot 1\\ &=5\frac{127}8\end{align}$$

Answer 1

You asked specifically about your mistake. The glaring error is that $\displaystyle \frac{128}{2} = 64$, not $16$ (in the cancellation step).

But this may have been a typo, because your final answer is actually correct. $\displaystyle 5(\frac{127}{8}) = \frac{635}{8}$.

Answer 2

Consider \begin{align} S_{n} = \sum_{k=1}^{n} x^{k-1} \end{align} which can be seen as the following. \begin{align} S_{n} &= \sum_{k=1}^{n} x^{k-1} + \sum_{k=n+1}^{\infty} x^{k-1} - \sum_{k=n+1}^{\infty} x^{k-1} \\ &= \sum_{k=1}^{\infty} x^{k-1} - \sum_{k=0}^{\infty} x^{n+k} \\ &= \frac{1}{1-x} - \frac{x^{n}}{1-x} = \frac{1-x^{n}}{1-x}. \end{align} Now, when $x=1/2$ this becomes \begin{align} \sum_{k=1}^{n} \left(\frac{1}{2}\right)^{k-1} = 2-2^{1-n} = \frac{2^{n+1}-2}{2^{n}}. \end{align} when $n=7$ this is further reduced to \begin{align} \sum_{k=1}^{7} \left(\frac{1}{2}\right)^{k-1} = \frac{127}{64}. \end{align} Multiplying both sides by 40 yields the answer desired, namely, \begin{align} 40 \sum_{k=1}^{7} \left(\frac{1}{2}\right)^{k-1} = \frac{635}{8}. \end{align}

Answer 3

$$\sum_{k=1}^{7}40 \left( \frac{1}{2}\right)^{k-1} = \frac{635}{8}$$

Take the left hand side and factor out constants:

$$\displaystyle \sum\limits_{k=1}^{7}40 \left( \frac{1}{2}\right)^{k-1} = 20 \sum\limits_{k=1}^7 \left(\frac 12\right)^k$$

Then use $\displaystyle\sum\limits_{k=1}^n r^k = \frac{r(1-r^n)}{1-r}$, where $r=\frac 12, n=7$