Question

A sequence is defined as t1=1 and
t(n+1)=tn+3. which is the nth term of the
sequence ?
A) 3n-2
B) 4n-3
C) n^2-n+1
D) 5n-4
E) 6n-5


I need this for a test :(( in t1, the 1 is a subscript and in t(n+1) the (n+1) is a subscript.

Answer 1

The recursive rule tells you that

$t_2=t_1+3$

$t_3=t_2+3=(t_1+3)+3=t_1+2\cdot3$

$t_4=t_3+3=(t_1+2\cdot3)+3=t_1+3\cdot3$

and so on, with the general rule

$t_n=t_1+(n-1)\cdot3$

Then with $t_1=1$, you have

$t_n=1+3(n-1)\implies\boxed{t_n=3n-2}$