Question

1280 - 640 + 320 -...+ 5find the sum of the series

Answer 1

Answer:

855

Explanation:

Given the series:

$1280-640+320-\cdots+5$

From observation, the series is geometric.

$\begin{gathered} -\frac{640}{1280}=-\frac{1}{2} \\ \frac{320}{-640}=-\frac{1}{2} \end{gathered}$

• The first term of the series, a = 1280

,

• The common ratio, r =-1/2

Since the series is finite, we find the number of terms in the series using the formula for the nth term of a geometric series:

$\begin{gathered} U_n=ar^{n-1} \\ 5=1280\left(-\frac{1}{2}\right)^{n-1} \\ \frac{5}{1280}=\left(-\frac{1}{2}\right)^{n-1} \\ \frac{1}{256}=\left(-\frac{1}{2}\right)^{n-1} \\ \frac{1}{2^8}=\left(-\frac{1}{2}\right)^{n-1} \\ 2^{-8}=(-2)^{-1(n-1)} \\ \text{ Assume n is odd} \\ 2^{-8}=(2)^{-(n-1)} \\ -8=-n+1 \\ n=1+8 \\ n=9 \end{gathered}$

This means that there are 9 terms in the series.

For a geometric series with a common ratio of less than 1, the sum is calculated using the formula:

$S_n=\frac{a(1-r^n)}{1-r}$

Substitute the values: a=1280, r=-1/2 and n=9

$\begin{gathered} S_9=\frac{1280\left(1-\left(-\frac{1}{2}\right)^9\right)}{1-\left(-\frac{1}{2}\right)} \\ =\frac{1280\left(1-\left(-\frac{1}{512}\right)\right)}{1+\frac{1}{2}} \\ =\frac{1280\left(1+\frac{1}{512}\right)}{1+\frac{1}{2}} \\ =855 \end{gathered}$

The sum of the series is 855.