Question

An infinite geometric series has S=}" alt="\frac{64}{3}" align="absmiddle" class="latex-formula"> 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 $S_k$:

$(1 - r)S_k = a\left(1 - r^k\right) \implies S_k = a\dfrac{1-r^k}{1-r}$

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 :

$\dfrac a{1-r} = \dfrac{64}3 \implies a = \dfrac{64(1-r)}3$

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 :

$\dfrac{64(1-r)}3 \left(1+r+r^2\right) = 21 \\\\ \dfrac{64}3 (1 - r^3) = 21 \implies r^3 = \dfrac1{64} \implies r = \dfrac14$

Now solve for a :

$a\left(1 + \dfrac14 + \dfrac1{4^2}\right) = 21 \implies a = 16$

It follows that

$S_5 = a\left(1 + r + r^2 + r^3 + r^4\right) \\\\ S_5 = 16\left(1 + \dfrac14 + \dfrac1{16} + \dfrac1{64} + \dfrac1{256}\right) = \boxed{\frac{341}{16}} = 21 + \dfrac5{16} = 21.3125$