SOLUTION: I'm wondering: what is the sixth term in this sequence: -1, 0, 0, 2, 10, ?? Thank you!

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Question 1062558: I'm wondering: what is the sixth term in this sequence: -1, 0, 0, 2, 10, ??
Thank you!
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

General formula:

a%5Bn%5D%22%22=%22%22%28n-2%29%28n-3%29%28n%5E2%2B7n-20%29%2F24

a%5B1%5D%22%22=%22%22%281-2%29%281-3%29%281%5E2%2B7%2A1-20%29%2F24%22%22=%22%22%28-1%29%28-2%29%281%2B7-20%29%2F24%22%22=%22%22%28-1%29%28-2%29%28-12%29%2F24%22%22=%22%22%28-24%29%2F24%22%22=%22%22-1

a%5B2%5D%22%22=%22%22%282-2%29%282-3%29%282%5E2%2B7%2A2-20%29%2F24%22%22=%22%22%280%29%28-1%29%284%2B49-20%29%2F24%22%22=%22%220

a%5B3%5D%22%22=%22%22%283-2%29%283-3%29%283%5E2%2B7%2A3-20%29%2F24%22%22=%22%22%281%29%280%29%281%2B7-20%29%2F24%22%22=%22%220

a%5B4%5D%22%22=%22%22%284-2%29%284-3%29%284%5E2%2B7%2A4-20%29%2F24%22%22=%22%22%282%29%281%29%2816%2B28-20%29%2F24%22%22=%22%22%282%29%281%29%2824%29%2F24%22%22=%22%22%2848%29%2F24%22%22=%22%222

a%5B5%5D%22%22=%22%22%285-2%29%285-3%29%285%5E2%2B7%2A5-20%29%2F24%22%22=%22%22%283%29%282%29%2825%2B35-20%29%2F24%22%22=%22%22%283%29%282%29%2840%29%2F24%22%22=%22%22%28240%29%2F24%22%22=%22%2210

So the next term:

a%5B6%5D%22%22=%22%22%286-2%29%286-3%29%286%5E2%2B7%2A6-20%29%2F24%22%22=%22%22%284%29%283%29%2836%2B42-20%29%2F24%22%22=%22%22%284%29%283%29%2858%29%2F24%22%22=%22%22%28696%29%2F24%22%22=%22%2229

Edwin

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.
I'm wondering: what is the sixth term in this sequence: -1, 0, 0, 2, 10, ??
Thank you!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Also, you can add the polynomial 

A*(n-1)*(n-2)*(n-3)*(n-4)*(n-5) 

with an arbitrary coefficient A (even not necessary integer !) to the polynomial discovered by Edwin, with the same susses. 


You can even to select the coefficient "A" to get ANY real number as a next term !


Actually, the correct answer to your question is: highlight%28ANY_NUMBER%29.