# How do you find the first five terms of the sequence a_1=3, a_(n+1)=2a_n−1?

How do you find the first five terms of the sequence ${a}_{1}=3$, ${a}_{n+1}=2{a}_{n}-1$?
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Jamari Morgan
${a}_{n+1}=2{a}_{n}-1$
So if ${a}_{n}=3$ then ${a}_{n+1}$ =. 2 x 3 - 1. = 5

an+1 is the next term in the sequence after an
And ${a}_{n}=5$ then ${a}_{n+1}$ =. 2 x 5 - 1 = 9
${a}_{n}=9$ then ${a}_{n+1}$ =2 x 9 - 1 = 17
And the next term 2 x 17 - 1 = 33

The sequence increases by powers of 2!