1. ## Sequence

How would I do this problem?

Use the formula a[sub-1]+a[sub-1]r+a[sub-1]r[squared]+...+a[sub-1]r[n-1 power]=

a[sub-1]* 1-r[nth power]/1-r

for the nth partial sum of a geometric sequence to find the sum 1+2/3+4/9+8/27+...+64/729.

701/2187
2059/729
2187/701
2187/729

2. Originally Posted by gretchen
How would I do this problem?

Use the formula a[sub-1]+a[sub-1]r+a[sub-1]r[squared]+...+a[sub-1]r[n-1 power]=

a[sub-1]* 1-r[nth power]/1-r

for the nth partial sum of a geometric sequence to find the sum 1+2/3+4/9+8/27+...+64/729.

701/2187
2059/729
2187/701
2187/729
The summation formula for a finite geometric series is:

a+a r+a r^2+...+a r^(n-1) = a [1-r^n]/[1-r]

in the example given a=1, r=(2/3), and n=7 (as 64/729=(2/3)^6).

So the sum of the given series is:

1 [1-(2/3)^7]/[1-2/3]=2059/729

RonL