The problem involves simplifying the expression (2n+4 - 2 x 2n) / (2n+2 x 4), with the expected answer being 7/8. The subject area relates to algebraic manipulation and simplification of expressions.
The discussion is ongoing, with participants providing insights and corrections regarding the expression's interpretation. Some have offered guidance on how to approach the simplification, while others are still questioning the initial setup and assumptions.
There is a noted emphasis on the importance of parentheses in the expression, as the lack of them has led to confusion and multiple interpretations. Participants are also addressing potential sign errors in the original problem statement.
[itex] <br /> Thanks for replying and I've added some brackets (the question didn't have any). Wouldn't the numerator be [itex]2^{n+4}- 2^{n+1}[/itex]?[/itex]HallsofIvy said:Please us parentheses! Is this
[itex](2^{n+4} - 2^{n+ 1}) / (2^{n+2}(4))[/itex]
or [itex]2^{n+4}- (2^{n+1}/2^{n+2})(4)[/itex]
or [itex]2^{n+4}- (2^{n+1})/(2^{n+2}(4))[/itex]?
Assuming it is the first, yes, the denominator will be [itex]2^{n+ 4}[/itex]. The numerator is [itex]2^{n+4}+ 2^{n+1}[/itex] 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 /> [tex]\frac{7(2^{n+1})}{2^{n+4}}[/tex][/itex]
Added a closing tex tag, and changed a leading itex tag to a tex tag.HallsofIvy said:Please us parentheses! Is this
[itex](2^{n+4} - 2^{n+ 1}) / (2^{n+2}(4))[/itex]
or [itex]2^{n+4}- (2^{n+1}/2^{n+2})(4)[/itex]
or [itex]2^{n+4}- (2^{n+1})/(2^{n+2}(4))[/itex]?
Sign error above that you corrected below. The numerator is 2n+4 - 2n+1.HallsofIvy said:Assuming it is the first, yes, the denominator will be [itex]2^{n+ 4}[/itex]. The numerator is [itex]2^{n+4}+ 2^{n+1}[/itex]
HallsofIvy said:which we can write as [tex]2^{n+1}2^3- 2^{n+1}= 8(2^{n+1})- 2^{n+1}= 2^{n+1}(8- 1)= 2^{n+1}(7).[/tex]
Can you reduce
[tex]\frac{7(2^{n+1})}{2^{n+4}}[/tex]