Question

What is the sum of the first 10 terms of this geometric sequence 3, –6, 12, –24, …?

Answer 1

`3, -6, 12, -24, ...`

You can rewrite this as:

`(-1)^n 3*(1, 2, 4, 8, ...)`

If we focus on `1, 2, 4, 8...`

`a_(n+1) = 2a_n`
with `a_0 = 1`.

This can be written out as a nice sum:
`sum_(n=0)^(N) 2^n = 2^0 + 2^1 + 2^2 + 2^3 + ...`
`= 1 + 2 + 4 + 8 + ...`

Thus, now we can recombine everything to get:

`color(green)(3sum_(n=0)^(N) (-1)^n 2^n)`

where `N` is some arbitrary stop point.

The formula for summing this is:

![http://www.mathsisfun.com/]()

`r = -2`
(since `(-1)^n * 2^n = (-2)^n`)

`a = 3`
`n = 10`

`=> 3((1-(-2)^10))/(1+2)` `-> (-2)^10 = 2^10`

`= 3((1-2^(5*2))/3)`

`= cancel(3)(1-32^2)/cancel(3) = 1-1024 = color(blue)(-1023)`

To sum this up more easily than brute-forcing it, if you don't remember the formula, there's something rather creative you can do.

If you look at the odd-indexed terms (`-6, -24, -96, -384,` etc), you can see that they are all successively multiplied by `4`. You can see that the even terms (`3, 12, 48, 192,` etc) are also this way. Thus, it lines up like so:

`n = 0 " " " 2 " " " 4 " " " " 6`:

`" " 3 + 12 + 48 + 192 + ...`

`n = 1 " " " 3 " " " 5 " " " " 7`:

`-6 -24 -96 -384 - ...`

and you always subtract twice the value of the positive number. Thus, just add up the first five even-indexed numbers (indices `0, 2, 4, ...`) and switch the sign.

`= -(3 + 12 + 48 + 192 + 768)`
`= -3(4^n)`
with `n_0 = 0`

`= -3(4^n) = -3[4^0 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5]`
`= -3(1 + 4 + 16 + 64 + 256)`
`= -3(341) = color(blue)(-1023)`

(You didn't have to write down as many terms here.)

Or, try adding up the original to check:
`= 3 - 6 + 12 - 24 + 48 - 96 + 192 - 384 + 768 - 1536)`
`= -3 - 12 - 48 - 192 - 768`
`= -15 - 240 - 768`
`= -255 - 768`
`= -223 - 800 = color(blue)(-1023)`

(Except now you had to write out 10 terms to add up.)