# How do you find the number of terms in the following geometric series: 1 + 2 + 4 + ... + 67108864?

Feb 13, 2016

Identify the appropriate power of $2$ to find that there are $27$ terms.

#### Explanation:

The general form of a term of a geometric sequence or series is:

${a}_{n} = a {r}^{n - 1}$

where $a$ is the first term and $r$ is the common ratio.

In our example, $a = 1$ and $r = 2$, so the question boils down to identifying which power of $2$ is $67108864$.

Notice that ${2}^{10} = 1024 \approx 1000 = {10}^{3}$ and $67108864$ is a little over $64 \cdot {10}^{6}$ hence we find the correct power is:

${2}^{26} = {2}^{6} \cdot {2}^{10} \cdot {2}^{10} = 64 \cdot 1024 \cdot 1024$

So there are $27$ terms: ${2}^{0} , {2}^{1} , {2}^{2} , \ldots , {2}^{26}$

Footnote

As a child, I used to like to write powers of $2$ on a blackboard, starting with $1$ and doubling it repeatedly.

In later life I found it useful to memorise powers of $2$ up to about ${2}^{32} = 4294967296$.

A couple of 'fun' ones are ${2}^{25} = 33554432$ and ${2}^{29} = 536870912$ (which contains all the digits except $4$).

Feb 13, 2016

$n = 27$

#### Explanation:

It was calculated that the ${n}^{t h}$ term in a geometric sequence or series is
${a}_{n} = a {r}^{n - 1}$
where $a$ is the first term and $r$ is the common ratio.
In the given question, $a = 1 \mathmr{and} r = 2$. Therefore

(To take it further from where @George C. left in his solution)

$67108864 = 1 \cdot {2}^{n - 1}$
$\implies 67108864 = {2}^{n - 1}$

To find $n$, taking log of both sides
$\log 67108864 = \log {2}^{n - 1}$
=> log 67108864=(n-1)log 2^
$\implies \left(n - 1\right) = \log \frac{67108864}{\log} 2$
$\implies \left(n - 1\right) = \frac{7.82678}{0.30103}$
$\implies \left(n - 1\right) = 26$
or $n = 27$