May 2, 2018

1)

${a}_{1} = 1$
${a}_{2} = {1}^{2} + 1 = 2$
${a}_{3} = {2}^{2} + 1 = 5$
${a}_{4} = {3}^{2} + 1 = 10$
${a}_{5} = {4}^{2} + 1 = 17$

2)

${a}_{1} = {1}^{2} + 2 \times 1 = 3$
${a}_{2} = {2}^{2} + 2 \times 2 = 8$
${a}_{3} = {3}^{2} + 2 \times 3 = 15$
${a}_{4} = {4}^{2} + 2 \times 4 = 24$
${a}_{5} = {5}^{2} + 2 \times 5 = 35$

May 2, 2018

$\text{see explanation}$

#### Explanation:

$\left(1\right)$

$\text{this is a "color(blue)"recurrence relationship}$

$\text{where each term is obtained by substituting the }$
$\text{previous term into the equation}$

$\text{the first term is given, that is } {a}_{1} = \textcolor{red}{1}$

$\Rightarrow {a}_{2} = {\left(\textcolor{red}{1}\right)}^{2} + 1 = 1 + 1 = \textcolor{b l u e}{2}$

$\Rightarrow {a}_{3} = {\left(\textcolor{b l u e}{2}\right)}^{2} + 1 = 4 + 1 = \textcolor{m a \ge n t a}{5}$

$\Rightarrow {a}_{4} = {\left(\textcolor{m a \ge n t a}{5}\right)}^{2} + 1 = 25 + 1 = \textcolor{p u r p \le}{26}$

${a}_{5} = {\left(\textcolor{p u r p \le}{26}\right)}^{2} + 1 = 676 + 1 = 677$

$\text{the first 5 terms are } 1 , 2 , 5 , 26 , 677$

$\left(2\right)$

$\text{obtain the terms in this sequence by substituting}$
$\text{1,2,3,4 and 5 into the n th term formula}$

$\Rightarrow {a}_{1} = {1}^{2} - \left(2 \times 1\right) = 1 - 2 = - 1$

${a}_{2} = {2}^{2} - \left(2 \times 2\right) = 4 - 4 = 0$

${a}_{3} = {3}^{2} - \left(2 \times 3\right) = 9 - 6 = 3$

${a}_{4} = {4}^{2} - \left(2 \times 4\right) = 16 - 8 = 8$

${a}_{5} = {5}^{2} - \left(2 \times 5\right) = 25 - 10 = 15$

$\text{the first 5 terms are } - 1 , 0 , 3 , 8 , 15$

May 2, 2018

$\textcolor{red}{\left(1\right)} 1 , 2 , 5 , 26 , 677.$
$\textcolor{red}{\left(2\right)} - 1 , 0 , 3 , 8 , 15.$

#### Explanation:

$\textcolor{b l u e}{\left(1\right) {a}_{1} = 1} , \mathmr{and}$

color(red)(a_n=(a_(n-1))^2+1

${a}_{2} = {\left({a}_{2 - 1}\right)}^{2} + 1 = {\left({a}_{1}\right)}^{2} + 1 = {\left(1\right)}^{2} + 1 = 1 + 1 = 2$

${a}_{3} = {\left({a}_{3 - 1}\right)}^{2} + 1 = {\left({a}_{2}\right)}^{2} + 1 = {\left(2\right)}^{2} + 1 = 4 + 1 = 5$

${a}_{4} = {\left({a}_{4 - 1}\right)}^{2} + 1 = {\left({a}_{3}\right)}^{2} + 1 = {\left(5\right)}^{2} + 1 = 25 + 1 = 26$

${a}_{5} = {\left({a}_{5 - 1}\right)}^{2} + 1 = {\left({a}_{4}\right)}^{2} + 1 = {\left(26\right)}^{2} + 1 = 676 + 1 = 677$

Hence. first five terms :$1 , 2 , 5 , 26 , 677.$

$\left(2\right)$

${a}_{n} = {n}^{2} - 2 n$

${a}_{1} = {\left(1\right)}^{2} - 2 \left(1\right) = 1 - 2 = - 1$

${a}_{2} = {\left(2\right)}^{2} - 2 \left(2\right) = 4 - 4 = 0$

${a}_{3} = {\left(3\right)}^{2} - 2 \left(3\right) = 9 - 6 = 3$

${a}_{4} = {\left(4\right)}^{2} - 2 \left(4\right) = 16 - 8 = 8$

${a}_{5} = {\left(5\right)}^{2} - 2 \left(5\right) = 25 - 10 = 15$

Hence. first five terms : $- 1 , 0 , 3 , 8 , 15.$