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

2 Answers
Sep 30, 2016

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!

Sep 30, 2016

"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.