explicit formula for recurrence relation

the recurrence relation is dsubk = 2dsub(k-1) + 3, for all integers k >= 2
dsub1 = 2

I need to find an explicit formula, and so far I have:
dsubn = 2^n + 3*2^(n-2) + 3*2^(n-3) + ... + 3*2^2 + 3*2 + 3
= 2^n + 3*(2^(k-1) - 1)/(2 - 1)   by factoring and sum of geometric sequence
= 2^n + 3*2^(k-1) - 3   algebra

now apparently dsubn = 5*2^(n-1) - 3, how do I get there?

Your working is correct, just factor out 2^(n-1) from 2^n + 3*2^(n-1) and the desired result trivially follows.

OP here, how did I get so far in my academic career and still can't figure out basic shit like this? Thanks.

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