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

2 Answers
Feb 13, 2016

Answer:

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 ~~ 1000 = 10^3# and #67108864# is a little over #64 * 10^6# hence we find the correct power is:

#2^26 = 2^6 * 2^10 * 2^10 = 64*1024*1024#

So there are #27# terms: #2^0, 2^1, 2^2,...,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

Answer:

#n=27#

Explanation:

It was calculated that the #n^(th)# term in a geometric sequence or series is
#a_n=ar^(n-1)#
where #a# is the first term and #r# is the common ratio.
In the given question, #a=1 and r=2#. Therefore

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

#67108864=1*2^(n-1)#
#=>67108864=2^(n-1)#

To find #n#, taking log of both sides
#log 67108864=log 2^(n-1)#
#=> log 67108864=(n-1)log 2^#
#=> (n-1)=log 67108864/log 2#
#=> (n-1)=7.82678/0.30103#
#=> (n-1)=26#
or #n=27#