Math Problem Solver

$$-8,-7,-5,-2,...$$The next 3 terms in this quadratic sequence are $a_{5}=2$ , $a_{6}=7$ , $a_{7}=13$.

The nth term in this sequence is given by the formula:

Explicit Formula: $$a_n=\frac{n^{2}}{2} - \frac{n}{2} - 8$$

Recursive Formula: $$a_n=a_{n-1} + n - 1, \text{where } a_{1}=-8$$

$$\begin{matrix}-8&&-7&&-5&&-2\\[6pt]&+1&&+2&&+3\\[6pt]&&+1&&+1\\[6pt]\end{matrix}$$

Explicit Formula

Since there are 2 rows of differences, the formula for the sequence can be written as a polynomial with degree 2, where $n$ is the term number and $(x_{0}, x_{1}, x_{2})$ are the coefficients: $$a_n=n^{2} x_{2} + n x_{1} + x_{0}$$

Using the first 3 terms in the sequence, create and solve the system of equations for $(x_{0}, x_{1}, x_{2})$: $$\begin{aligned} -8 &= 1^{2} x_{2} + 1 x_{1} + x_{0} \\ -7 &= 2^{2} x_{2} + 2 x_{1} + x_{0} \\ -5 &= 3^{2} x_{2} + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + x_{1} + x_{2} = -8\\x_{0} + 2 x_{1} + 4 x_{2} = -7\\x_{0} + 3 x_{1} + 9 x_{2} = -5 \end{aligned}$$ $$ \Rightarrow \quad (x_{0}, x_{1}, x_{2})=\left( -8, \ - \frac{1}{2}, \ \frac{1}{2}\right) $$

The n^th term rule is:$$\begin{aligned} a_n&=n^{2} x_{2} + n x_{1} + x_{0} \\ &=n^{2} \left(\frac{1}{2}\right) + n \left(- \frac{1}{2}\right) + \left(-8\right) \\ &=\frac{n^{2}}{2} - \frac{n}{2} - 8 \end{aligned}$$

Recursive Formula

Since there are 2 rows of differences, the formula for the sequence can be written as the sum of $a_{n-1}$ and polynomial with degree 1, where $n$ is the term number and $(x_{0}, x_{1})$ are the coefficients: $$a_n=a_{n-1} + n x_{1} + x_{0}$$

Using the first 3 terms in the sequence, create and solve the system of equations for $(x_{0}, x_{1})$: $$\begin{aligned} -7 &= -8 + 2 x_{1} + x_{0} \\ -5 &= -7 + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + 2 x_{1} = 1\\x_{0} + 3 x_{1} = 2 \end{aligned}$$ $$ \Rightarrow \quad (x_{0}, x_{1})=\left( -1, \ 1\right) $$

The n^th term rule is:$$\begin{aligned} a_n&=a_{n-1} + n x_{1} + x_{0} \\ &=a_{n-1} + n \left(1\right) + \left(-1\right) \\ &=a_{n-1} + n - 1 \end{aligned}$$