Find the first four partial sums and the nth partial sum of the sequence an. An=2/3^n

Answer 1

To find the partial sums of the sequence an = 2/3^n, we start by writing out the terms of the sequence:

a1 = 2/3^1 = 2/3
a2 = 2/3^2 = 2/9
a3 = 2/3^3 = 2/27
a4 = 2/3^4 = 2/81

Now, we can calculate the first four partial sums:

S1 = a1 = 2/3
S2 = a1 + a2 = 2/3 + 2/9 = 6/9 + 2/9 = 8/9
S3 = a1 + a2 + a3 = 2/3 + 2/9 + 2/27 = 18/27 + 6/27 + 2/27 = 26/27
S4 = a1 + a2 + a3 + a4 = 2/3 + 2/9 + 2/27 + 2/81 = 54/81 + 18/81 + 6/81 + 2/81 = 80/81

Therefore, the first four partial sums are:

S1 = 2/3
S2 = 8/9
S3 = 26/27
S4 = 80/81

The nth partial sum of the sequence can be calculated using the formula for the sum of a geometric series:

Sn = a1(1 - r^n) / (1 - r)

In this case, a1 = 2/3 and r = 1/3. Therefore, the nth partial sum of the sequence an = 2/3^n is:

Sn = (2/3)(1 - (1/3)^n) / (1 - 1/3)
Sn = (2/3)(1 - 1/3^n) / (2/3)
Sn = 1 - 1/3^n