Question

How do you find the sum of the terms of geometric 1, -2, 4, -8..... -8192?

Answer 1

`-5461`

Explanation:

The general term of a geometric sequence is given by the formula:

`a_n = a * r^(n-1)`

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

In our example `a = 1` and `r = -2`

If we want to sum the first `N` terms, we find:

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

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

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

`=sum_(n=2)^(N+1) a r^(n-1) - sum_(n=1)^N a r^(n-1)`

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

`=a(r^N-1)`

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

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

That's a nice general formula, but note also that:

`sum_(n=1)^N a_n = (a(r^N-1))/(r-1) = (ar^N-a)/(r-1) = (a_(N+1)-a_1)/(r-1)`

So rather than having to count the terms between `1` and `-8192`, we can just use the next term in the sequence, `-8192*-2 = 16364`.

Then:

`sum_(n=1)^N a_n = (a_(N+1)-a_1)/(r-1) = (16384-1)/(-2-1) = -16383/3 = -5461`