Baten80 wrote:
\(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?\)
A. \(2^9\)
B. \(2^{10}\)
C. \(2^{16}\)
D. \(2^{35}\)
E. \(2^{37}\)
don't neglect what might be the easiest way to solve this problem -- and what's absolutely the most straightforward: just convert these to concrete numbers.
the problem statement is
2 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256
the first few terms here are pretty insignificant, as far as the total sum is concerned. 32 + 64 is pretty close to 100, so, this sum will be close to 100 + 128 + 256.
the answer choices are
A/ 512
B/ 1024
C/ ha, you must be kidding
D/ ha, you must be kidding
E/ ha, you must be kidding.
this method certainly doesn't take any MORE time than any of the other methods here... and it may in fact take less, for many people.
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