How do you find the explicit formula for the following sequence 4,8,16,32,...?

2 Answers
May 2, 2016

Answer:

#a_n = 4*2^(n-1)# if it is a geometric sequence.

or could be: #a_n=1/3(2n^3-6n^2+16n)# if not.

Explanation:

There is a common ratio between successive pairs of terms:

#8/4 = 2#

#16/8 = 2#

#32/16 = 2#

So this looks like a geometric sequence with initial term #a=4# and common ratio #r=2#.

If so, the formula for the #n#th term is:

#a_n = a r^(n-1) = 4*2^(n-1)#

This is probably the answer expected by the questioner.

#color(white)()#
However, note that any finite sequence of terms does not determine an infinite sequence - unless you are told what kind of sequence it is - e.g. arithmetic, geometric, harmonic.

For example, we can match these first #4# terms with a cubic formula as follows:

Write down the sequence as a list:

#color(blue)(4), 8, 16, 32#

Write down the sequence of differences between each pair of terms:

#color(purple)(4), 8, 16#

Write down the sequence of differences of this sequence:

#color(magenta)(4), 8#

Write down the sequence of differences of this sequence:

#color(red)(4)#

Having reached a constant sequence (albeit consisting of only one term), we can use the initial term of each of the sequences we have found as coefficients of a formula for the #n#th term:

#a_n = color(blue)(4)/(0!) + color(purple)(4)/(1!)(n-1) + color(magenta)(4)/(2!)(n-1)(n-2) + color(red)(4)/(3!)(n-1)(n-2)(n-3)#

#=4+(4n-4)+(2n^2-6n+4)+(2/3n^3-4n^2+22/3n-4)#

#=2/3n^3-2n^2+16/3n#

#=1/3(2n^3-6n^2+16n)#

May 2, 2016

Answer:

#T_n = 2^(n+1)#

Explanation:

I am not sure of the use of the word explicit, but perhaps this will help.

4, 8, 16, 32 are all powers of 2, but the sequence does not start from the first power of 2 which would be 2.

#n# tells us the position in the sequence of a particular term.
#n# always starts from 1 for the first term, 2 for the second and so on.. Try to find a pattern or rule which links #n# to the value of the corresponding term.

We would like to have #2^1 = 2#, #2^2 = 4#, #2^3=8# as the first three terms, because then the index #n# matches the position in the sequence.

However in the sequence we are working with, the first term is 4, which is 2², with the index being 1 more than #n#.

The terms can then be given as #2^(n+1)#

The general term would be #T_n = 2^(n+1)#