Math Problem Solver

$$14,27,53,92,144,...$$The next term in this quadratic sequence is $a_{6}=209$.

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

Explicit Formula: $$a_n=\frac{13 n^{2}}{2} - \frac{13 n}{2} + 14$$

Recursive Formula: $$a_n=a_{n-1} + 13 n - 13, \text{where } a_{1}=14$$

$$\begin{matrix}14&&27&&53&&92&&144\\[6pt]&+13&&+26&&+39&&+52\\[6pt]&&+13&&+13&&+13\\[6pt]\end{matrix}$$

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 $(x_{0}, x_{1}, x_{2})$: $$\begin{aligned} 14 &= 1^{2} x_{2} + 1 x_{1} + x_{0} \\ 27 &= 2^{2} x_{2} + 2 x_{1} + x_{0} \\ 53 &= 3^{2} x_{2} + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + x_{1} + x_{2} = 14\\x_{0} + 2 x_{1} + 4 x_{2} = 27\\x_{0} + 3 x_{1} + 9 x_{2} = 53 \end{aligned}$$ $$ \Rightarrow \quad (x_{0}, x_{1}, x_{2})=\left( 14, \ - \frac{13}{2}, \ \frac{13}{2}\right) $$

The n^th term rule is:$$\begin{aligned} a_n&=n^{2} x_{2} + n x_{1} + x_{0} \\ &=n^{2} \left(\frac{13}{2}\right) + n \left(- \frac{13}{2}\right) + \left(14\right) \\ &=\frac{13 n^{2}}{2} - \frac{13 n}{2} + 14 \end{aligned}$$

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 $(x_{0}, x_{1})$: $$\begin{aligned} 27 &= 14 + 2 x_{1} + x_{0} \\ 53 &= 27 + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + 2 x_{1} = 13\\x_{0} + 3 x_{1} = 26 \end{aligned}$$ $$ \Rightarrow \quad (x_{0}, x_{1})=\left( -13, \ 13\right) $$

The n^th term rule is:$$\begin{aligned} a_n&=a_{n-1} + n x_{1} + x_{0} \\ &=a_{n-1} + n \left(13\right) + \left(-13\right) \\ &=a_{n-1} + 13 n - 13 \end{aligned}$$