Math Problem Solver

$$-2,2,10,22,38,...$$This infinite sequence is a quadratic sequence. The n^th term rule for the sequence is $a_n=2 n^{2} - 2 n - 2$, where $n>0$. The recursive formula is $a_n=a_{n-1} + 4 n - 4$, where $n>1$ and $a_{1}=-2$.

$$\begin{matrix}-2&&2&&10&&22&&38\\[6pt]&+4&&+8&&+12&&+16\\[6pt]&&+4&&+4&&+4\\[6pt]\end{matrix}$$

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 $(x_{0}, x_{1}, x_{2})$: $$\begin{aligned} -2 &= 1^{2} x_{2} + 1 x_{1} + x_{0} \\ 2 &= 2^{2} x_{2} + 2 x_{1} + x_{0} \\ 10 &= 3^{2} x_{2} + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + x_{1} + x_{2} = -2\\x_{0} + 2 x_{1} + 4 x_{2} = 2\\x_{0} + 3 x_{1} + 9 x_{2} = 10 \end{aligned}$$ $$ \Rightarrow \quad (x_{0}, x_{1}, x_{2})=\left( -2, \ -2, \ 2\right) $$

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

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 $(x_{0}, x_{1})$: $$\begin{aligned} 2 &= -2 + 2 x_{1} + x_{0} \\ 10 &= 2 + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + 2 x_{1} = 4\\x_{0} + 3 x_{1} = 8 \end{aligned}$$ $$ \Rightarrow \quad (x_{0}, x_{1})=\left( -4, \ 4\right) $$

The n^th term rule is:$$\begin{aligned} a_n&=a_{n-1} + n x_{1} + x_{0} \\ &=a_{n-1} + n \left(4\right) + \left(-4\right) \\ &=a_{n-1} + 4 n - 4 \end{aligned}$$