Simplify Homework: (2n+4 - 2 x 2n) / (2n+2 x 4) | 7/8

[dh743]

Homework Statement

Simplify: (2ⁿ⁺⁴ - 2 x 2ⁿ) / (2ⁿ⁺² x 4)

Homework Equations

N/A

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

 

[HallsofIvy]

Please use parentheses! Is this
$(2^{n+4} - 2^{n+ 1}) / (2^{n+2}(4))$
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

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

 

[dh743]

Please us parentheses! Is this
$(2^{n+4} - 2^{n+ 1}) / (2^{n+2}(4))$
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 [itex]2^{n+1}2^3- 2^{n+1}= 8(2^{n+1})- 2^{n+1}= 2^{n+1}(8- 1)= 2^{n+1}(7).<br /> <br /> Can you reduce <br /> <br /> $$\frac{7(2^{n+1})}{2^{n+4}}$$[/itex]

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

 

[Mark44]

Please us parentheses! Is this
$(2^{n+4} - 2^{n+ 1}) / (2^{n+2}(4))$
or $2^{n+4}- (2^{n+1}/2^{n+2})(4)$
or $2^{n+4}- (2^{n+1})/(2^{n+2}(4))$?

Added a closing tex tag, and changed a leading itex tag to a tex tag.

Assuming it is the first, yes, the denominator will be $2^{n+ 4}$. The numerator is $2^{n+4}+ 2^{n+1}$

Sign error above that you corrected below. The numerator is 2ⁿ⁺⁴ - 2ⁿ⁺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

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

 

[dh743]

Thank you it makes sense now.