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December 20, 2014

December 20, 2014

Posted by **Tk** on Monday, March 21, 2011 at 6:02am.

- arithmetic -
**Reiny**, Monday, March 21, 2011 at 8:58amthe general term of a quadratic sequence is

term(n) = an^2 + bn + c, where a, b, and c are constants

It can be shown using differences that a is (1/2) of the second difference.

Thus a = 1

so term(1) = 8

8 = 1(1^2) + b(1) + c ---> b+c = 7

term(4) = 32

32 = 1(2^2) + b(2) + c ---> 2b + c = 28

subtract them: b = 21

then c = 7-21 = -14

term(n) = n^2 + 21n - 14

- correction - arithmetic -
**Reiny**, Monday, March 21, 2011 at 9:07amI used the 32 as if it were the 2nd terms, should have been the 4th

Here is the correct version

Thus a = 1

so term(1) = 8

8 = 1(1^2) + b(1) + c ---> b+c = 7

term(4) = 32

32 = 1(4^2) + b(4) + c ---> 4b + c = 16

subtract them:

3b = 9

b = 3

then c = 7-3 = 4

term(n) = n^2 + 3n + 4

check:

t(1) = 1+3+4 = 8

t(4) = 16 + 12 + 4 = 32

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