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Archangel Macsika

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

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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


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