Question

How do you find the sum of `12(-1/2)^(i-1)` from i=0 to 9?

Answer 1

The is a geometric series that has `9 - 0 + 1 = 10` terms.

The common ratio will be `-1/2`, and the first term will be `12(-1/2)^(-1) = 12(-2)^1 = -24`.

We use the formula `s_n = (a(1 - r^n))/(1 - r), r !=1` to find the sum of a geometric series.

`s_10 = (-24(1 - (-1/2)^10))/(1 - (-1/2))`

`s_10= -15.984375~= -15.98`

Hopefully this helps!

Answer 2

`"The reqd. Sum="-1023/64.`

Explanation:

Reqd. Sum`=sum_0^9 {12(-1/2)^(i-1)}=12sum_0^9(-1/2)^i(-1/2)^-1}`

`=12sum_0^9(-1/2)^i(-2)}=-24sum_0^9(-1/2)^i`

`=-24[(-1/2)^0+(-1/2)^1+(-1/2)^2+... ...+(-1/2)^9}`

`=-24{1+(-1/2)^1+(-1/2)^2+... ...+(-1/2)^9}`

We realise that the series in `{... ...}` is a Geometric Series,

having `"the First Term, "a=1", and, the Common Ratio "r=-1/2,`

and no. of terms `n=10`

For such a series, the sum `S_n,` of the first `n` terms is,

`S_n=(a(1-r^n))/(1-r).` Therefore,

`S_10=((1)(1-(-1/2)^10))/((1-(-1/2)))=(1-1/1024)/(1+1/2)=1023/1024*2/3`

Finally, we get,

`"The reqd. Sum="-24*1023/64*2/3=-1023/64.`