In a quadratic sequence, when finding the nth term the formula is an2xbn+c...
a=1/2 the 2nd constant difference
c=the 0th term
and b?????????
I've been trying to solve this for 2 years and the maths teacher that taught me this method left, no one else seems to know :(
Please help, any other methods accepted too.
2006-10-23 02:34:35 · 4 answers · asked by stardom 2 in Science & Mathematics ➔ Mathematics
For every n we have a(n+1) - a(n) = a [(n+1)^2 - n^2] + b(n+1 - n) = a(2n +1) + b . If we make n =0, then we get a(1) - a(0) = a(1) - c = a +b so that b = a(1) - c + a. a(1) is the term of order 1. So, b is the sum of a and the differfence between the termos of order 1 and 0.
2006-10-23 03:41:02 · answer #1 · answered by Steiner 7 · 0⤊ 0⤋
attempt subtracting n² from each and each term and see what sequence think approximately: 4, 12, 24, 40, 60 attempt subtracting yet another n² 3, 8, 15, 24, 35 Subtract yet another n² 2, 4, 6, 8, 10 There we flow. a pretty effective sequence. So we subtracted 3n² and are left with 2n So, the nth term is: 3n² + 2n = n(3n + 2) (be conscious - it is very a tribulation and mistake approach. i do no longer comprehend if it is going to artwork for all quadratic sequences.)
2016-10-16 07:22:30 · answer #2 · answered by ? 4 · 0⤊ 0⤋
i haven't a clue it appears to work on multiples of four
2006-10-23 02:41:42 · answer #3 · answered by beegeecee 2 · 0⤊ 2⤋
i dont understnd ur question.
2006-10-23 03:32:57 · answer #4 · answered by sumone^^ 3 · 0⤊ 1⤋