What is the nth term of #2#, #1#, #3/2#, #2/1#,... ?
1 Answer
One matching formula is:
#a_n = 1/4(-n^3+9n^2-24n+24)#
but no sequence is determined purely by its first few terms.
Explanation:
No finite sequence determines the following numbers, unless you have additional information about the sequence (e.g. that it is arithmetic, geometric, quadratic, etc.).
That having been said, we can find a matching polynomial formula by analysing the differences between successive terms:
Write down the given sequence:
#color(blue)(2), 1, 3/2, 2/1#
Write down the sequence of differences between consecutive terms:
#color(blue)(-1), 1/2, 1/2#
Write down the sequence of differences between those differences:
#color(blue)(3/2), 0#
Write down the sequence of differences of those differences:
#color(blue)(-3/2)#
Having reached a constant sequence (albeit of just one term), we can use the initial term of each of these sequences as coefficients to write down a formula:
#a_n = color(blue)(2)/(0!)+color(blue)(-1)/(1!)(n-1)+color(blue)(3/2)/(2!)(n-1)(n-2)+color(blue)(-3/2)/(3!)(n-1)(n-2)(n-3)#
#color(white)(a_n) = 2-n+1+3/4n^2-9/4n+3/2-1/4n^3+3/2n^2-11/4n+3/2#
#color(white)(a_n) = -1/4n^3+9/4n^2-6n+6#
#color(white)(a_n) = 1/4(-n^3+9n^2-24n+24)#
