The expression for the nth term of the quadratic sequence is:
T^n =n^2 −8n+3
A quadratic sequence is one where the second finite difference is constant. Let's find the common difference and the second finite difference for the given sequence.
The given sequence is: -6, -11, -14, -15, -14
First differences: -11 - (-6) = -5, -14 - (-11) = -3, -15 - (-14) = -1, -14 - (-15) = 1
Second differences: -3 - (-5) = 2, -1 - (-3) = 2, 1 - (-1) = 2
Since the second finite difference is constant (in this case, 2), we can write the quadratic expression for the nth term of the sequence.
Let a be the coefficient of n^2 in the quadratic expression. The second finite difference is 2a, so 2a=2, which implies a=1.
Now, we can write the quadratic expression:
T^n =an^2 +bn+c
Since a=1, the expression becomes:
T^n =n^2 +bn+c
Now, we need to find the values of b and c. We can use the first two terms of the sequence to create two equations.
When n=1:
T^1 =1+b+c=−6
When n=2:
T^2 =4+2b+c=−11
Solving these equations simultaneously will give us the values of
b and
1+b+c=−6
4+2b+c=−11
Subtracting equation (1) from equation (2):
3+b=−5
Therefore, b=−8.
Now, substitute b=−8 into equation (1) to find
1+(−8)+c=−6
Solving for c, we get c=3.
So, the expression for the nth term of the quadratic sequence is:
T^n =n^2 −8n+3