Powers of 2

Relevant wiki: Rules of Exponents - Algebraic

This expression can be simplified as

$\large\dfrac{2^n(1 + 2^{-1})}{2^n(2^1 - 1)}$

= $\large\dfrac{1+\dfrac{1}{2}}{1}$

= $\large\boxed{\dfrac{3}{2}}$

Relevant wiki: Rules of Exponents

$\dfrac{2^n+2^{n-1}}{2^{n+1}-2^n}=\dfrac{2^n+\frac{2^n}{2}}{2^n(2)-2^n}=\dfrac{2^n(1+\frac{1}{2})}{2^n(2-1)}=\dfrac{\frac{3}{2}}{1}=\dfrac{3}{2}=$ $\huge\boxed{\color{#D61F06}1.5}$ $\color{#3D99F6}\large \boxed{answer}$

By subbing in 1 for n or any number for that matter we will get 3/2

Let $2^n$ =P

$\frac{P+P/2}{2P-P}$

= $\frac{2P+P}{2(2P-P)}$

= $\frac{2P+P}{4P-2P}$

= $\frac{3P}{2P}$

Cut P AND P

= $\frac{3}{2}$

We know that $2^{n+1}-2^n=2^n$ . Thus $\dfrac{2^n+2^{n-1}}{2^{n+1}-2^n}=\dfrac{2^n+2^{n-1}}{2^n}=1+\dfrac12=\dfrac32=1.5$

The correct answer is 1.5 of course, but unusually the problem as posed says n is a <real> number, not the usual <integer>. This shouldn't make any difference, but I guess if you got really sneaky and started mixing up the +ve/-ve sign of sqrt(2) (n=0.5) you could conceivably get something else. By letting n=0.333... you could even make it complex, but you'd be doing some socially unacceptable things with the maths.

Taking any no. in the place of n

Try with n=1, and you're done... Sure, in that way you cannot be sure the expression is going to simplify "every time" to 3/2, but taking that for granted... This would have been more an interesting question in case the author would have chosen to make an "input a value" type question.

Take 2 raise to power n common