Question

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

Answer 1

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!

Answer 2

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

or:

$a_n=n^{2+1} \\ a_1= 1^{2+1} =1^3=1 \\ a_2=2^{2+1}=2^3=8 \\ a_3=3^{2+1} =3^3=27 \\ a_4=4^{2+1} =4^3=64 \\ a_5=5^{2+1} =5^3=125 \\ a_6=6^{2+1} =6^3=216$