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Here are the first five terms of a quadratic sequence, -6,-11,-14,-15,-14 find an expression,in terms of n, for the nth term of this sequence

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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

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