Question

An infinite geometric series has S=
`(64)/(3)` and
`S_(3)`=21. Find
`S_(5)`.

Answer 1

Since 64/3 = 21 + 1/3 > 21, I assume S is supposed to be the value of the infinite sum. So we have for some constants a and r (where |r | < 1),

`S = \displaystyle \sum_(n=1)^\infty ar^(n-1) = \frac{64}3 \\\\ S_3 = \sum_(n=1)^3 ar^(n-1) = 21`

Consider the k-th partial sum of the series,

`S_k = \displaystyle \sum_(n=1)^k ar^(n-1) = a \left(1 + r + r^2 + \cdots + r^(k-1)\right)`

Multiply both sides by r :

`rS_k = a\left(r + r^2 + r^3 + \cdots + r^k\right)`

Subtract this from
`S_k`:

`(1 - r)S_k = a\left(1 - r^k\right) \implies S_k = a(1-r^k)/(1-r)`

Now as k goes to ∞, the r ᵏ term converges to 0, which leaves us with

`S = \displaystyle \lim_(k\to\infty)S_k = \frac a{1-r} = \frac{64}3`

which we can solve for a :

`\frac a{1-r} = \frac{64}3 \implies a = \frac{64(1-r)}3`

Meanwhile, the 3rd partial sum is given to be

`\displaystyle S_3 = \sum_(k=1)^3 ar^(n-1) = a\left(1+r+r^2\right) = 21`

Substitute a into this equation and solve for r :

`\frac{64(1-r)}3 \left(1+r+r^2\right) = 21 \\\\ \frac{64}3 (1 - r^3) = 21 \implies r^3 = \frac1{64} \implies r = \frac14`

Now solve for a :

`a\left(1 + \frac14 + \frac1{4^2}\right) = 21 \implies a = 16`

It follows that

`S_5 = a\left(1 + r + r^2 + r^3 + r^4\right) \\\\ S_5 = 16\left(1 + \frac14 + \frac1{16} + \frac1{64} + \frac1{256}\right) = \boxed{(341)/(16)} = 21 + \frac5{16} = 21.3125`