Question

How do you find the sum of the geometric series `4096-512+64-...` to 5 terms?

Answer 1

3641

Explanation:

First get the common ratio by dividing and of the terms by its preceding term, like second term being divided by first, or third term being divided by 2nd. the common ratio is `-1/8`

Sum of a geometric series is given by the formula `S_n = a (1-r^n )/(1-r)` , when ( r<1)

Here a= 4096, `r=-1/8` and n=5

Accordingly, `S_5= 4096 (1- (-1/8)^5)/ (1-(-1/8)`

=`4096 (1+ 1/32768)/(9/8)`

=`4096 (32769)/32768 8/9`= 3641

Answer 2

`4096-512+64-8+1 = 3641`

Explanation:

Given:

`4096-512+64-...`

We are told that this is a geometric series, so there should be a common ratio between successive terms.

Note that:

`(-512)/4096 = -1/8`

`64/(-512) = -1/8`

So there is a common ratio of `-1/8`.

So the fourth and fifth terms must be:

`64*(-1/8) = -8`

and

`(-8)*(-1/8) = 1`

So the sum to `5` terms is:

`4096-512+64-8+1 = 3641`

`color(white)()`
The general term of a geometric series can be written:

`a_n = ar^(n-1)`

where `a` is the initial term and `r` the common ratio.

Then we find:

`(1-r) sum_(n=1)^N a_n = sum_(n=1)^N ar^(n-1) - r sum_(n=1)^N ar^(n-1)`

`color(white)((1-r) sum_(n=1)^N a_n) = sum_(n=1)^N ar^(n-1) - sum_(n=2)^(N+1) ar^(n-1)`

`color(white)((1-r) sum_(n=1)^N a_n) = a+color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - ar^N`

`color(white)((1-r) sum_(n=1)^N a_n) = a(1-r^N)`

So dividing both ends by `(1-r)` we find:

`sum_(n=1)^N a_n = (a(1-r^N))/(1-r)`

In our example we have `N=5`, `a=4096`, `r=-1/8`

So:

`sum_(n=1)^5 a_n = (4096(1-(-1/8)^5))/(1-(-1/8))`

`color(white)(sum_(n=1)^5 a_n) = (4096(1+1/32768))/(9/8)`

`color(white)(sum_(n=1)^5 a_n) = (4096*32769)/(9*4096)`

`color(white)(sum_(n=1)^5 a_n) = 32769/9`

`color(white)(sum_(n=1)^5 a_n) = 3641`

`color(white)()`
Bonus

If, in addition (as in our example) we have `abs(r) < 1`, then `lim_(N->oo) r^N = 0` and hence:

`sum_(n=1)^oo a_n = lim_(N->oo) (a(1-r^N))/(1-r) = a/(1-r)`

So in our example, the sum to infinity would be:

`4096/(1-(-1/8)) = 4096/(9/8) = 32768/9 = 3640.bar(8)`