Math Problem Solver

$$99,94,85,72,...$$This infinite sequence is a quadratic sequence. The n^th term rule for the sequence is $a_n=- 2 n^{2} + n + 100$, where $n>0$. The recursive formula is $a_n=a_{n-1} - 4 n + 3$, where $n>1$ and $a_{1}=99$.

$$\begin{matrix}99&&94&&85&&72\\[6pt]&-5&&-9&&-13\\[6pt]&&-4&&-4\\[6pt]\end{matrix}$$

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 $(x_{0}, x_{1}, x_{2})$: $$\begin{aligned} 99 &= 1^{2} x_{2} + 1 x_{1} + x_{0} \\ 94 &= 2^{2} x_{2} + 2 x_{1} + x_{0} \\ 85 &= 3^{2} x_{2} + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + x_{1} + x_{2} = 99\\x_{0} + 2 x_{1} + 4 x_{2} = 94\\x_{0} + 3 x_{1} + 9 x_{2} = 85 \end{aligned}$$ $$ \Rightarrow \quad (x_{0}, x_{1}, x_{2})=\left( 100, \ 1, \ -2\right) $$

The n^th term rule is:$$\begin{aligned} a_n&=n^{2} x_{2} + n x_{1} + x_{0} \\ &=n^{2} \left(-2\right) + n \left(1\right) + \left(100\right) \\ &=- 2 n^{2} + n + 100 \end{aligned}$$

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 $(x_{0}, x_{1})$: $$\begin{aligned} 94 &= 99 + 2 x_{1} + x_{0} \\ 85 &= 94 + 3 x_{1} + x_{0} \end{aligned} \quad \Rightarrow \quad \begin{aligned} x_{0} + 2 x_{1} = -5\\x_{0} + 3 x_{1} = -9 \end{aligned}$$ $$ \Rightarrow \quad (x_{0}, x_{1})=\left( 3, \ -4\right) $$

The n^th term rule is:$$\begin{aligned} a_n&=a_{n-1} + n x_{1} + x_{0} \\ &=a_{n-1} + n \left(-4\right) + \left(3\right) \\ &=a_{n-1} - 4 n + 3 \end{aligned}$$