Question 419106: 2,8,18,32,50 find the formula for the nth term
Found 2 solutions by Theo, acubed1234:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula that i came up with is this:
a[n] = 6 + 6*n + (sum(4*(n-1)) from n = 0 to n = n)
it works like this:
a[0] = 6 + 0 + (-4) = 2
a[1] = 6 + 6 + (-4 + 0) = 8
a[2] = 6 + 12 + (-4 + 0 + 4) = 18
a[3] = 6 + 18 + (-4 + 0 + 4 + 8) = 32
a[4] = 6 + 24 + (-4 + 0 + 4 + 8 + 12) = 50
it works for all of the numbers listed.
n has to be greater than or equal to 0
when n = 0, 4 * (n-1) becomes 4 * (-1) = -4
when n = 1, 4 * (n-1) becomes 4 * (0) = 0
when n = 2, 4 * (n-1) becomes 4 * (1) = 4
when n = 3, 4 * (n-1) becomes 4 * (2) = 8
when n = 4, 4 * (n-1) becomes 4 * (3) = 12
Answer by acubed1234(5)  (Show Source):
You can put this solution on YOUR website! 2n^2 gives
2,8,18,32,50,72, . . . .
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However,
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n^5-15n^4+85n^3-223n^2+274n-120 gives
2,8,18,32,50,192, . . . .
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But then there is,
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(2/5)n^5-6n^4+34n^3-88n^2+(548/5)n-48 that gives
2,8,18,32,50,120, . . . .
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And then there is ....
... I can keep going all day, but I'll just stop there.
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All these work, take your pick :)
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Finally, the point also is that given any finite set of numbers, the set does not necessarily define any single sequence.
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