Question

Give a recursive definition of the sequence $\left(a_n\right)\mathbf{,}\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{\dots }\mathbf{.}\mathbf{.}$

$\begin{array}{r}{\mathbf{a}}_{\mathbf{n}}\mathbf{=}\mathbf{4}\mathbf{n}\mathbf{-}\mathbf{2}\\ {\mathbf{a}}_{\mathbf{n}}\mathbf{=}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\\ {\mathbf{a}}_{\mathbf{n}}\mathbf{=}\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\\ {\mathbf{a}}_{\mathbf{n}}\mathbf{=}{\mathbf{n}}^{\mathbf{2}}\end{array}$

Short answer

$$\begin{gathered} \left(\mathrm{a}\right){\mathrm{a}}_1=2\mathrm{and}{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}+4\mathrm{when}\mathrm{n}\ge 2 \\ \left(\mathrm{b}\right){\mathrm{a}}_1=0\mathrm{and}{\mathrm{a}}_2=2\mathrm{and}{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-2}\mathrm{when}\mathrm{n}\ge 3 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{c}\right){\mathrm{a}}_1=2\mathrm{and}{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}+2n\mathrm{when}\mathrm{n}\ge 2 \\ \left(\mathrm{b}\right){\mathrm{a}}_1=1\mathrm{and}{\mathrm{a}}_n=a_{n-1}+2_{\mathrm{n}-1}\mathrm{when}\mathrm{n}\ge 2 \end{gathered}$$

Step-by-step solution

The recursive definition of the sequence:

A sequence can also be defined recursively, meaning that the previous terms define successive terms in the sequence. The recursive sequence is obtained by the deriving each successive term such that it is 2 larger than the previous obtained term.

To find recursive definition of the given sequence:  an=4n-2(a)

It is given that: $a_n=4n-2$

Determine the first value at n = 1

$a_1=4\left(1\right)-2=4-2=2$

Next determine the recursive definition by writing $a_n$in terms of $a_{n-1}$

$\begin{array}{r}{a}_n=4n-2\\ =4n-4+4-2\\ =\left[4\right(n-1)-2]+4\\ =a_{n-1}+4\end{array}$

Thus, the recursive definition is$a_1=2\text{, and}{a}_n=a_{n-1}+4\text{for}n\ge 2$

Step 3:To find recursive definition of the given sequence  an=1+(-1)n(b)

It is given that: $a_n=1+{\left(-1\right)}^n$

Determine the first two values at n = 1 and n = 2

$\begin{array}{r}{a}_1=1+(-1{)}^1=1-1=0\\ a_2=1+(-1{)}^2=1+1=2\end{array}$

Next determine the recursive definition by writing $a_n$in terms of $a_{n-2}$

$\begin{array}{r}{a}_n=1+(-1{)}^n\\ =1+(-1{)}^2\cdot (-1{)}^{n-2}\\ =1+1\cdot (-1{)}^{n-2}\\ =a_{n-2}\end{array}$

Thus, the recursive definition is$a_1=0,a_2=2,\text{and}{a}_n=a_{n-2}\text{for}n\ge 3$

To find recursive definition of the given sequence  an=n(n+1)   (c)

It is given that: $a_n=n\left(n+1\right)$

Determine the first value at n = 1

$a_1=1(1+1)=2$

Next determine the recursive definition by writing $a_n$in terms of $a_{n-1}$

$\begin{array}{r}{a}_n=n(n+1)\\ =n^2+n\\ =n^2+n-n+n\end{array}$

Also,

$\begin{array}{r}{a}_n=n(n-1)+n+n\\ =n(n-1)+2n\\ =a_{n-1}+2n\end{array}$

Thus, the recursive definition is ,$a_1=2\text{, and}{a}_n=a_{n-1}+2n\text{for}n\ge 2\text{.}$

To find recursive definition of the given sequence  an=n2 (d)

It is given that: $a_n=n^2$

Determine the first value at n = 1

$a_1=1^2=1$

Next determine the recursive definition by writing $a_n$in terms of $a_{n-1}$

$\begin{array}{r}{a}_n=n^2-2n+2n-1+1\\ =\left(n^2-2n+1\right)+2n-1\\ =(n-1{)}^2+2n-1\\ =a_{n-1}+2n-1\end{array}$

Thus, the recursive definition is $a_1=1,a_n=a_{n-1}+2n-1\text{and}n\ge 2\text{for}n\ge 2$,