Question

If the sum of 2^0 + 2^1 + 2^2 + ... + 2^k is at least 70,000, then the smallest possible value of k is

Answer 1

Since \(2^{16}\leq70000\leq2^{17}\), \(k=16\) would be the value you're looking for.

Answer 2

You might need to first prove that \[2^0+2^1+...+2^{k-1} < 2^k\]

Answer 3

If you're picky, my first post should read:

Since \(2^{16}<70000<2^{17}\), \(k=16\) would be the value you're looking for.

Answer 4

thnks i got it too. can u help me with another question: Determine the sum of the first 8 terms of a geometric series, given that t3 = 6 and the common ratio is -0.6. (Approximate the answer to 1 decimal place.)

Answer 5

is the answer 10.2?

Answer 6

I'm also getting 10.2

Answer 7

okay.. thanks

Answer 8

you're welcome