Question

An=n^2+1 first 6 terms of the sequence

Answer 1

`a_n=n^2+1 \\ \\ a_1=1^2+1=1+1=2 \\ a_2=2^2+1=4+1=5 \\ a_3=3^2+1=9+1=10 \\ a_4=4^2+1=16+1=17 \\ a_5=5^2+1=25+1=26 \\ a_6=6^2+1=36+1=37`

or:


`a_n=n^(2+1) \\ a_1= 1^(2+1) =1^3=1 \\ a_2=2^(2+1)=2^3=8 \\ a_3=3^(2+1) =3^3=27 \\ a_4=4^(2+1) =4^3=64 \\ a_5=5^(2+1) =5^3=125 \\ a_6=6^(2+1) =6^3=216`

Answer 2

Hello!

You put the numbers 1 to 6 in for n

=
`1^(2+1)= 1`
=
`2^(2+1)= 8`
=
`3^(2+1)= 27`
=
`4^(2+1)= 64`
=
`5^(2+1)= 125`
=
`6^(2+1)= 216`

The first 6 numbers are 1, 8, 27, 64, 125, 216

Hope this helps!