Question

Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions.

$\begin{array}{r}{\mathrm{a}}_{\mathrm{n}}=-2{\mathrm{a}}_{\mathrm{n}-1},{\mathrm{a}}_0=-1\\ {\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}-{\mathrm{a}}_{\mathrm{n}-2},{\mathrm{a}}_0=2,{\mathrm{a}}_1=-1\\ {\mathrm{a}}_{\mathrm{n}}=3{\mathrm{a}}_{\mathrm{n}-1}^2,{\mathrm{a}}_0=1\\ {\mathrm{a}}_{\mathrm{n}}={\mathrm{na}}_{\mathrm{n}-1}+{\mathrm{a}}_{\mathrm{n}-2}^2,{\mathrm{a}}_0=-1,{\mathrm{a}}_1=0\\ {\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}-{\mathrm{a}}_{\mathrm{n}-2}+{\mathrm{a}}_{\mathrm{n}-3},{\mathrm{a}}_0=1,{\mathrm{a}}_2=2\end{array}$

Short answer

Hence, the solution is

$\begin{array}{r}-1,2,-4,8,-16,32\\ 2,-1,-3,-2,1,3\\ 1,3,27,2187,14348907,617673396283947\\ -1,0,1,3,13,74\\ 1,1,2,2,1,1\end{array}$

Step-by-step solution

Step 1

A recurrence relation for the sequence $\left\{{\mathrm{a}}_{\mathrm{n}}\right\}$is an equation that expressesin terms of one or more of the previous terms of the sequence, namely, ${\mathrm{a}}_0,{\mathrm{a}}_1,.....,{\mathrm{a}}_{\mathrm{n}-1}$, for all integers n > ${\mathrm{n}}_0$with, where${\mathrm{n}}_0$ is a nonnegative integer. A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation.

Step 2

a ) Given

$$\begin{gathered} {\mathrm{a}}_{\mathrm{n}}=-2{\mathrm{a}}_{\mathrm{n}-1} \\ {\mathrm{a}}_0=-1 \end{gathered}$$

The first term is -1 .each term is the previous term multiplied by - 2
$\begin{array}{r}{\mathrm{a}}_0=-1\\ {\mathrm{a}}_1=-2{\mathrm{a}}_1=-2(-1)=2\\ {\mathrm{a}}_2=-2{\mathrm{a}}_1=-2\left(2\right)=4\\ {\mathrm{a}}_3=-2{\mathrm{a}}_2=-2\left(4\right)=8\\ {\mathrm{a}}_4=-2{\mathrm{a}}_3=-2\left(8\right)=-16\\ {\mathrm{a}}_5=-2{\mathrm{a}}_4=-2(-16)=32\end{array}$

Step 3

b) Given

$\begin{array}{r}{a}_n=a_{n-1}-a_{n-2}\\ a_0=2\\ a_1=-1\end{array}$

The first term isand the second term is - 1 . Each term is difference of the previous two terms.

$\begin{array}{r}{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}-{\mathrm{a}}_{\mathrm{n}-2}\\ {\mathrm{a}}_0=2\\ {\mathrm{a}}_1=-1\\ {\mathrm{a}}_2={\mathrm{a}}_1-{\mathrm{a}}_0=-1-2=-3\\ {\mathrm{a}}_3={\mathrm{a}}_2-{\mathrm{a}}_1=-3-(-1)=-2\\ {\mathrm{a}}_4={\mathrm{a}}_3-{\mathrm{a}}_2=-2-(-3)=1\\ {\mathrm{a}}_5={\mathrm{a}}_4-{\mathrm{a}}_3=1-(-2)=3\end{array}$

Step 4

c) Given

$$\begin{gathered} {\mathrm{a}}_{\mathrm{n}}=3{{\mathrm{a}}^2}_{\mathrm{n}-1} \\ {\mathrm{a}}_0=1 \end{gathered}$$

The first term is.each term is the square of the previous term multiplied by 3

$\begin{array}{r}{\mathrm{a}}_0=1\\ {\mathrm{a}}_1=3{\mathrm{a}}_0^2=3\left(1^2\right)=3\\ {\mathrm{a}}_2=3{\mathrm{a}}_1^2=3\left(3^2\right)=3\left(9\right)=3\\ {\mathrm{a}}_3=3{\mathrm{a}}_2^2=3\left({27}^2\right)=3\left(729\right)=2187\\ {\mathrm{a}}_4=3{\mathrm{a}}_3^2=3\left({2187}^2\right)=3\left(4782969\right)=14348907\\ {\mathrm{a}}_5=3{\mathrm{a}}_4^2=3\left({14348907}^2\right)=3\left(205891132094649\right)=617673396283947\end{array}$

Step 5

d) Given

$\begin{array}{r}{a}_n=n{a}_{n-1}+a_{n-2}^2\\ a_0=-1\\ a_1=0\end{array}$

The first term is 2 and the second term is -1 the nth term is multiplied by the previous term and increased by the square of the term before the previous term.

$\begin{array}{r}{\mathrm{a}}_0=-1\\ {\mathrm{a}}_1=0\\ {\mathrm{a}}_2=2{\mathrm{a}}_1+{\mathrm{a}}_0^2=2\left(0\right)+(-1{)}^2=0+1=1\\ {\mathrm{a}}_3=3{\mathrm{a}}_2+{\mathrm{a}}_1^2=3\left(1\right)+(0{)}^2=3+0=3\\ {\mathrm{a}}_4=4{\mathrm{a}}_3+{\mathrm{a}}_2^2=4\left(3\right)+(1{)}^2=12+1=13\\ {\mathrm{a}}_5=5{\mathrm{a}}_4+{\mathrm{a}}_3^2=5\left(13\right)+(3{)}^2=65+9=74\end{array}$

Step 6:

e) Given

$\begin{array}{r}{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}-{\mathrm{a}}_{\mathrm{n}-2}+{\mathrm{a}}_{\mathrm{n}-3}\\ {\mathrm{a}}_0=1\\ {\mathrm{a}}_1=1\\ {\mathrm{a}}_2=2\end{array}$

The first term is, the second term is 1 and the third term is 2 . Each term is difference of the two previous terms increased by the third previous term.

$\begin{array}{r}{\mathrm{a}}_0=1\\ {\mathrm{a}}_1=1\\ {\mathrm{a}}_2=2\\ {\mathrm{a}}_3={\mathrm{a}}_2-{\mathrm{a}}_1+{\mathrm{a}}_0=2-1+1=2\\ {\mathrm{a}}_4={\mathrm{a}}_3-{\mathrm{a}}_2+{\mathrm{a}}_1=2-2+1=1\\ {\mathrm{a}}_5={\mathrm{a}}_4-{\mathrm{a}}_3+{\mathrm{a}}_2=1-2+2=1\end{array}$

Hence, the first six terms of the sequence

$\begin{array}{r}-1,2,-4,8,-16,32\\ 2,-1,-3,-2,1,3\\ 1,3,27,2187,14348907,617673396283947\\ -1,0,1,3,13,74\\ 1,1,2,2,1,1\end{array}$