Question

The sum to infinity of a geometric series is 32.The sum of the first four terms is 30 and

Answer 1

The terms in the series are

$\displaystyle \sum_{n=1}^\infty ar^{n-1} = a + ar + ar^2 + ar^3 + \cdots$

The series converges, so we $|r|$. All the terms in this particular series are known to be positive, so $0Consider the [tex]N$-th partial sum of the series,

$S_N = \displaystyle \sum_{n=1}^N ar^{n-1} = a + ar + ar^2 + \cdots + ar^{N-1}$

Multiply both sides by $r$ :

$r S_N = ar + ar^2 + ar^3 + \cdots + ar^N$

Subtracting this from $S_N$ eliminates all the $r^n$ terms between $n=1$ and $n=N-1$, leaving us with

$(1 - r) S_N = a - ar^N \implies S_N = \dfrac{a(1-r^N)}{1-r}$

As $N\to\infty$, the $r^N$ term will converge to zero, and the value of the infinite series is

$\displaystyle \sum_{n=1}^\infty ar^{n-1} = \lim_{n\to\infty} S_N = S = \frac a{1-r}$

Now, we're given

$S = \dfrac a{1-r} = 32 ~~~~ \text{ and } ~~~~ S_4 = \dfrac{a(1-r^4)}{1-r} = 30$

By substitution, we have

$S_4 = 32(1-r^4) = 30 \implies 1-r^4 = \dfrac{15}{16} \implies r^4 = \dfrac1{16} \implies r = \dfrac12$

so that

$S = \dfrac a{1 - \frac12} = 32 \implies a = 16$

Then the sum of the first 8 terms is

$S_8 = \dfrac{a(1-r^8)}{1-r} = \dfrac{16\left(1 - \frac1{2^8}\right)}{1-\frac12} = \dfrac{255}8$

and the difference between this and the infinite sum is

$S - S_8 = 32 - \dfrac{255}8 = \dfrac{256-255}8 = \boxed{\dfrac18}$