Question

The 100th term of 1, 7, 13, 19, … is

Answer 1

`Consider \: the \: sequence \: u_(n) \\ u_(n) \: appears \: to \: be \: arithmtic`

Quick Proof:

`u_(4)-u_(3)=u_(3)-u_(2)=... \\ 19 - 13 = 13 - 7 = .. \\ 6 = 6 = ...`

`u_(n) \: is \: arithmetic \: with \: d = 6 \:and \: u_(1) = 1 \\`

`u_(n) = u_(1)+(n-1)d \\ u_(100) = 1 + (100 - 1)6 \\ u_(100) = 595`