Find (showing your reasoning) a formula for the term tn of the sequence defined by t1 = 4; tn = 2tn−1, n ≥ 2.

$t_n=2t_{n-1},n\ge 2$

Then

$r=\dfrac{t_n}{t_{n-1}}=\dfrac{2t_{n-1}}{t_{n-1}}=2, n\geq2$

This geometric progression with the initial term $t=t_1=4$ and the common ratio $r=2$

$4, 4(2), 4(2)^2,...$

The nth term of this geometric progression is

$t_n=tr^{n-1}$

$t_n=4(2)^{n-1},n\geq1$