What is the formula for the #n#th term of the sequence #1, 3, 9, 19,...# ?
1 Answer
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#
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
#a_n = color(blue)(1) + color(blue)(2)(n-1)#
To match the third term too, we need to add a multiple of
#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