Question

Let `a_n` be a sequence given by: `{1, 6, 15, 28, 45,66,..., f(n)}`. Show that the generating function `f(n)` is of the form `an^2 + bn + c`. Find the formula by computing the coefficients `a, b, c`?

Answer 1

`:. P_n^6 = 2n^2-n`

Explanation:

Strategy:
Take the given sequence find the difference between consecutive numbers:
`P_n = {1,6,15,28,45,66, 91,120,cdots}`
Step 1 `rArr` Layer 1
`{1,5,9,13,17,21, cdots}`
Step 2 `rArr` Layer 2 , Do it again
`{4, 4, 4, 4, 4,cdots}`
Taking the difference is in discrete math is the same as taking the derivative (i.e. slope). took two subtraction (two layers) before we reached a comstant number `4`, that means the sequence is polynomial growth.
Give that I asert that: `P_n = an^2+bn+c`
All I have to do now find the value of `a, b and c`
To solve for `a,b and c` I use the first 3 entry of the sequence setting `n = {1,2,3}`

`Eq.1 rArr` `P_1 = a+b+c = 1`
`Eq.2 rArr` `P_2 = 4a+2b+c = 6`
`Eq.3 rArr` `P_3 = 9a+3b+c = 15`
`[[1,1,1], [4,2,1], [9,3,1]]xx[[a],[b],[c]] = [[1],[6],[15]]`

Solving a, b, c using any matrix calculator on the internet:
`[[a],[b],[c]] = [[2],[-1],[0]]`
`:. P_n^6 = 2n^2-n`

Check: `P_1^6 = 1; P_2^6 = 6; P_3^6 = 15;` checks out

PS: You can also use python, I used python simply... It is cool