# Indice Problem

1. Aug 14, 2010

### dh743

1. The problem statement, all variables and given/known data
Simplify: (2n+4 - 2 x 2n) / (2n+2 x 4)

2. Relevant equations
N/A

3. The attempt at a solution
(2n+4 - 2n+1) / (2n+4)

Unsure where to go from here, but the given answer is 7/8
Thanks for any help

Last edited: Aug 14, 2010
2. Aug 14, 2010

### HallsofIvy

$(2^{n+4} - 2^{n+ 1}) / (2^{n+2}(4))$
or $2^{n+4}- (2^{n+1}/2^{n+2})(4)$
or $2^{n+4}- (2^{n+1})/(2^{n+2}(4))$?

Assuming it is the first, yes, the denominator will be $2^{n+ 4}$. The numerator is $2^{n+4}+ 2^{n+1}$ which we can write as $2^{n+1}2^3- 2^{n+1}=$$8(2^{n+1})- 2^{n+1}=$$2^{n+1}(8- 1)= 2^{n+1}(7)$.

Can you reduce

$$\frac{7(2^{n+1})}{2^{n+4}}$$

Last edited by a moderator: Aug 15, 2010
3. Aug 14, 2010

### dh743

Thanks for replying and I've added some brackets (the question didn't have any). Wouldn't the numerator be $2^{n+4}- 2^{n+1}$?

4. Aug 14, 2010

### Staff: Mentor

Added a closing tex tag, and changed a leading itex tag to a tex tag.
Sign error above that you corrected below. The numerator is 2n+4 - 2n+1.

5. Aug 14, 2010

### dh743

Thank you it makes sense now.