What is the formula for the #n#th term of the sequence #1, 3, 9, 19,...# ?

1 Answer
Feb 3, 2017

Answer:

#a_n = 2n^2-4n+3#

Explanation:

Let us examine the differences between successive pairs of terms.

Write down the given sequence:

#color(blue)(1), 3, 9, 19#

Write down the sequence of differences between successive pairs of terms:

#color(blue)(2), 6, 10#

Write down the sequence of difference between successive pairs of those terms:

#color(blue)(4), 4#

Having arrived at a constant sequence after two steps, we can use the first term of each of the above sequences as coefficients to write down a quadratic expression that matches the given sequence:

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

#color(white)(a_n) = 1+2n-2+2n^2-6n+4#

#color(white)(a_n) = 2n^2-4n+3#

#color(white)()#
Footnote

Here are some more details as to where the formula above comes from.

Given a sequence:

#1, 3, 9, 19#

we can try matching it term by term as follows:

If we only need to match the first term, then we could write:

#a_n = color(blue)(1)#

To match both the first and the second terms, we need to add a multiple of #(n-1)#, to avoid throwing away what we have achieved so far, but what multiple? It will be proportional to the difference between the first and second terms of the sequence we were given, that is #color(blue)(2)#. In addition note that #((2)-1) = 1#, so we do not need to divide the #color(blue)(2)# by anything to get our next formula:

#a_n = color(blue)(1) + color(blue)(2)(n-1)#

To match the third term too, we need to add a multiple of #(n-1)(n-2)#, to avoid throwing away what we have achieved so far. Note that #((3)-1)((3)-2) = 2*1 = 2#, so we need to divide the difference #color(blue)(4)# by #2# to get the correct multiplier for #(n-1)(n-2)# and our next formula:

#a_n = color(blue)(1) + color(blue)(2)(n-1) + color(blue)(4)/2(n-1)(n-2)#

That formula is sufficient to match the given quadratic sequence.

In general the multiplier for #(n-1)(n-2)...(n-k)# will be the number we find by taking differences, divided by #k!#.