9514 1404 393
Answer:
t(n) = -n^2 -n +102
Step-by-step explanation:
We can use the first 3 terms to find a, b, c.
a(1)^2 +b(1) +c = 100
a(2)^2 +b(2) +c = 96
a(3)^2 +b(3) +c = 90
Subtract the first equation from the other two:
3a +b = -4 . . . . . . [eq4]
8a +2b = -10 . . . . [eq5]
[eq5] can be reduced to
4a +b = -5 . . . . . . [eq6]
Subtracting [eq4] from [eq6] gives ...
(4a +b) -(3a +b) = (-5) -(-4)
a = -1 . . . . . . . . simplify
Filling that into [eq4], we get
3(-1) +b = -4
b = -1 . . . . . . . add 3
Putting the values for 'a' and 'b' into the first equation gives ...
-1 +-1 +c = 100
c = 102
Then the n-th term t(n) of the sequence is ...
t(n) = -n^2 -n +102
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<em>Additional comment</em>
As you might guess, there is a generic solution for quadratic sequences. In terms of the first term (a1), first first difference (d1), and second difference (d2), the coefficients can be written as ...
- a = d2/2
- b = d1 -3a
- c = a1 -d1 +d2
This sequence has first differences of -4, -6, -8, -10, and second differences of -2. Putting the sequence values into the above formulas, we find ...
- a = -2/2 = -1
- b = -4 -3(-1) = -1
- c = 100 -(-4) +(-2) = 102 . . . . . as above
