Question

Find the first five terms in sequence with the following a. 2n^2+6

Answer 1

the formula is:

`a_n=2\cdot n^2+6`

and we need to find a1, a2, a3, a4 and a5

n=1

`\begin{gathered} a_1=2\cdot1^2+6 \\ a_1=2+6=8 \end{gathered}`

n=2

`\begin{gathered} a_2=2\cdot2^2+6 \\ a_2=2\cdot4+6 \\ a_2=8+6=14 \end{gathered}`

n=3

`\begin{gathered} a_3=2\cdot3^2+6 \\ a_3=2\cdot9+6 \\ a_3=18+6=24 \end{gathered}`

n=4

`\begin{gathered} a_4=2\cdot4^2+6 \\ a_4=2\cdot16+6 \\ a_4=32+6=38 \end{gathered}`

n=5

`\begin{gathered} a_5=2\cdot5^2+6 \\ a_5=2\cdot25+6 \\ a_5=50+6=56 \end{gathered}`

so the answer is:

a1=8

a2=14

a3=24

a4=38

a5=56