Math Problem Solver

$$1,8,21,40,65,...$$This infinite sequence is a quadratic sequence. The n^th term rule for the sequence is $a_n=n \left(3 n - 2\right)$, where $n>0$. The recursive formula is $a_n=a_{n-1} + 6 n - 5$, where $n>1$ and $a_{1}=1$.

$$\begin{matrix}1&&8&&21&&40&&65\\[6pt]&+7&&+13&&+19&&+25\\[6pt]&&+6&&+6&&+6\\[6pt]\end{matrix}$$

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 $(x_{0}, x_{1}, x_{2})$: $$\begin{aligned} 1 &= 1^{2} x_{2} + 1 x_{1} + x_{0} \\ 8 &= 2^{2} x_{2} + 2 x_{1} + x_{0} \\ 21 &= 3^{2} x_{2} + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + x_{1} + x_{2} = 1\\x_{0} + 2 x_{1} + 4 x_{2} = 8\\x_{0} + 3 x_{1} + 9 x_{2} = 21 \end{aligned}$$ $$\Rightarrow \quad (x_{0}, x_{1}, x_{2})=\left( 0, \ -2, \ 3\right)$$

The n^th term rule is:$$\begin{aligned} a_n&=n^{2} x_{2} + n x_{1} + x_{0} \\ &=n^{2} \left(3\right) + n \left(-2\right) + \left(0\right) \\ &=n \left(3 n - 2\right) \end{aligned}$$

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 $(x_{0}, x_{1})$: $$\begin{aligned} 8 &= 1 + 2 x_{1} + x_{0} \\ 21 &= 8 + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + 2 x_{1} = 7\\x_{0} + 3 x_{1} = 13 \end{aligned}$$ $$\Rightarrow \quad (x_{0}, x_{1})=\left( -5, \ 6\right)$$

The n^th term rule is:$$\begin{aligned} a_n&=a_{n-1} + n x_{1} + x_{0} \\ &=a_{n-1} + n \left(6\right) + \left(-5\right) \\ &=a_{n-1} + 6 n - 5 \end{aligned}$$