Leaderboard wrote:

The question has a bug. You mention that Sn = 2^(n - 1), but from your description the first term is 2, which is 2^1, not 2^0. And also there isn't quite any correct answer either, because

S_n = a(r^n - 1)/(r - 1). Then S_n = 2(2^n - 1) or (2^n - 1) depending on what a is, and I don't think any of the options fit it.

When n=10

S = S_1 + S_2 + .... + S_9 + S_10

=> S = 2 + 2^1 + 2^2 + 2^3 + .... + 2^8 + 2^9

=> S - 2 = 2^1 + 2^2 + 2^3 + .... + 2^8 + 2^9

Lets consider RHS as X

X = 2^1 + 2^2 + 2^3 + .... + 2^8 + 2^9

2*X = 2^2 + 2^3 + .... + 2^8 + 2^9 + 2^10

subtract first equation from second

=> 2*X - X = -2^1 + 0 + 0 + 0 + .... + 0 + 0 + 2^10

=> X = 2^10 - 2

S = X + 2 = 2^10

The answer is B.

It would have been better if it was mentioned that S_n = 2^(n-1) for n>1.