What is the formula for the sequence #4,2,3,4,3,2,...# ?

1 Answer
Nov 5, 2016

Answer:

#a_n = 1/120 (3 n^5-40 n^4+145 n^3+40 n^2-868 n+1200)#

Explanation:

If you want a formula for a sequence with these as the first six terms then you can proceed as follows:

Write down the original sequence:

#color(blue)(4), 2, 3, 4, 3, 2#

Write down a sequence formed by the differences of pairs of consecutive terms:

#color(blue)(-2), 1, 1, -1, -1#

Write down a sequence formed by the differences of pairs of consecutive terms of that sequence:

#color(blue)(3), 0, -2, 0#

Write down the sequence of differences of that sequence:

#color(blue)(-3), -2, 2#

Write down the sequence of differences of that sequence:

#color(blue)(1), 4#

Write down the sequence of differences of that sequence:

#color(blue)(3)#

We can then use the initial terms of these sequences as the coefficients of a formula for the #n#th term:

#a_n = color(blue)(4)/(0!) + color(blue)(-2)/(1!)(n-1) + color(blue)(3)/(2!)(n-1)(n-2) + color(blue)(-3)/(3!)(n-1)(n-2)(n-3) + color(blue)(1)/(4!)(n-1)(n-2)(n-3)(n-4) + color(blue)(3)/(5!)(n-1)(n-2)(n-3)(n-4)(n-5)#

#color(white)(a_n) = 1/120 (3 n^5-40 n^4+145 n^3+40 n^2-868 n+1200)#

#color(white)()#
Footnote

Note carefully that I said a formula. A finite sequence does not determine a unique formula. The above formula is the simplest polynomial function that matches the given values.