9

Σ 3×2^n-1

n=1

S_9=

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9

Σ 3 ⋅ 2^(n – 1)

n = 1

sum_(n=1)^9(3 ⋅ 2^n – 1) = 3057

The formula I have is

n-1

Σ (ar^k) = a (1-r^n) / (1-r)

k=0

[I've listed my source below.]

This formula needs a bit of translation, and I need to make an assumption that you've been just a bit sloppy in transcribing the sequence we're summing. I think it ought to be

9

Σ 3×2^(n-1)

n=1

That is, the -1 in part of the exponent, because otherwise it's not (quite) simply a geometric expression.

But since the exponents in that expression actually run from 0 to 8,

we can translate it to the standard formula (as I have it)

with n=9, a=3, r=2 and conclude that

S_9 = 3 (1 – 2^9) / (1 – 2)

= 3 (-511) / (-1)

= 3 (511) = 1533