Question

For the geometric sequence 4/3, 2/3, 1/3, 1/6, find the sum of S

Answer 1

The formula for a geometric sequence is:

$a_n=a_1r^{(n-1)}$

where a_1 is the first term of a sequence and r is the "coomon ratio"

We can see that theach term is half the previous term, thus r = 1/2

We can write:

$a_n=\frac{4}{3}\cdot(\frac{1}{2})^{(n-1)}$

Now, that we know the sequence formula, we can write the sum of the sequence:

$\sum ^{n-1}_{k\mathop=0}(ar^k)=a(\frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}})$

Thus:

$\sum ^{n-1}_{k\mathop{=}0}(ar^k)=2a(1-(\frac{1}{2})^n)$