<p>If n is a positive integer and 2^n + 2^n+1 = k, what is 2^n+2 in terms of k?</p>
<p>A) k-1 / 2
B) 4k / 3
C) 2k
D) 2k + 1
E) k^2</p>
<p>Can someone go through the steps and tell me how the answer is B? do you add n and n+1 together? What do you do? I'm stuck!</p>
<p>first of all 2^n+1 should be written 2^(n+1) so we know that (n+1) is the entire exponent…</p>
<p>2^n means 2 times 2 times 2 until the 2 repeats n times
2^(n+1) means 2 times 2 times 2 until the 2 repeats (n+1) times</p>
<p>that means 2^(n+1) is basically 2^n except you multiply it by 2 ONE more time.
likewise, 2^(n+2) is basically 2^n except you multiply it by 2 TWO more times.</p>
<p>so you can write 2^(n+1) as 2<em>(2^n)
so you can write 2^(n+2) as 4</em>(2^n)
for comprehension sake, 2^(n+3) would be 8<em>(2^n), and 2^(n+4) would be 16</em>(2^n)</p>
<p>back to the problem:</p>
<p>(2^n) + 2^(n+1) = k
(2^n) + 2<em>(2^n) = k
3</em>(2^n) = k</p>
<p>2^(n+2) = 4(2^n)*</p>
<p>we need to find 4<em>(2^n) in terms of k, which is 3</em>(2^n)</p>
<p>multiply k, which is 3<em>(2^n), by 4/3 to get 4</em>(2^n)</p>
<p>4*(2^n) = 4k/3</p>
<p>You should put parentheses around your exponents. Otherwise, it changes the equation completely and makes it unsolvable.
2^n + 2^(n+1) = k
2^n + (2^n)(2)
2^n (1+2)
3(2^n) = k
2^n = k/3
(2^2)(2^n) = (2^2)(k/3)
2^(n+2) = 4k/3</p>
<p>edit: wow, i was late. :((</p>