Answer:
f(n) = -n^2 -3n +5
Explanation:
Suppose the formula is ...
f(n) = an^2 +bn +c
Then we have ...
f(1) = 1 = a(1^2) +b(1) +c
f(2) = -5 = a(2^2) +b(2) +c
f(3) = -13 = a(3^2) +b(3) +c
__
Here's a way to solve these equations.
Subtract the first equation from the second:
-6 = 3a +b . . . . . 4th equation
Subtract the second equation from the third:
-8 = 5a +b . . . . . 5th equation
Subtract the fourth equation from the fifth:
-2 = 2a
a = -1
Then substituting into the 4th equation to find b, we have ...
-6 = 3(-1) +b
-3 = b
and ...
1 = -1 +(-3) +c . . . . . substituting "a" and "b" into the first equation
5 = c
The formula is ...
f(n) = -n^2 -3n +5