Question

Write the first three terms of the following sequences whose $n^{th}$ terms are given by
(i) $a_n=\frac{n(n-2)}{3}$
(ii) $c_n=(-1{)}^n{3}^{n+2}$
(iii) $z_n=\frac{(-1{)}^{n}n(n+2)}{4}$

Answer 1

(i) The $n$th term of the given sequence is $a_n=\frac{n(n-2)}{3}$.

To find the first three terms of the given sequence, substitute $n=1,2$ and $3$ in $a_n=\frac{n(n-2)}{3}$ as shown below:

$a_1=\frac{1(1-2)}{3}=\frac{1\times -1}{3}=-\frac{1}{3}$

$a_2=\frac{2(2-2)}{3}=\frac{2\times 0}{3}=\frac{0}{3}=0$

$a_3=\frac{3(3-2)}{3}=\frac{3\times 1}{3}=\frac{3}{3}=1$

Hence the first three terms of the sequence $a_n=\frac{n(n-2)}{3}$ are $-\frac{1}{3},0$ and $1$.

(ii) The $n$th term of the given sequence is $c_n={(-1)}^n{3}^{n+2}$.

To find the first three terms of the given sequence, substitute $n=1,2$ and $3$ in $c_n={(-1)}^n{3}^{n+2}$ as shown below:

$c_1={(-1)}^1{3}^{1+2}=-1\times 3^3=-1\times 27=-27$

$c_2={(-1)}^2{3}^{2+2}=1\times 3^4=1\times 81=81$

$c_3={(-1)}^3{3}^{3+2}=-1\times 3^5=-1\times 243=-243$

Hence the first three terms of the sequence $c_n={(-1)}^n{3}^{n+2}$ are $-27,81$ and $-243$.

(iii) The $n$th term of the given sequence is $z_n=\frac{{(-1)}^{n}n(n+2)}{4}$.

To find the first three terms of the given sequence, substitute $n=1,2$ and $3$ in $z_n=\frac{{(-1)}^{n}n(n+2)}{4}$ as shown below:

$z_1=\frac{{(-1)}^{1}1(1+2)}{4}=\frac{-1\times 1\times 3}{4}=-\frac{3}{4}$

$z_2=\frac{{(-1)}^{2}2(2+2)}{4}=\frac{1\times 2\times 4}{4}=\frac{8}{4}=2$

$z_3=\frac{{(-1)}^{3}3(3+2)}{4}=\frac{-1\times 3\times 5}{4}=-\frac{15}{4}$

Hence the first three terms of the sequence $z_n=\frac{{(-1)}^{n}n(n+2)}{4}$ are $-\frac{3}{4},2$ and $-\frac{15}{4}$.