Simplify Top and Bottom First

Algebra Level 2

$\large \dfrac{(4^{100} + 2^{150})(4^{100}- 2^{150}) }{(2+2^2+2^3+\cdots+2^{100} )} = \, ?$

2^{150}

2^{199}

2^{100}

1

2^{299}

100

1 solution

J C
Mar 17, 2016

The numerator will come out to $(4^{100}+2^{150})(4^{100}-2^{150})=4^{200}-2^{300}=2^{400}-2^{300}=2^{300}(2^{100}-1)$

So $2^1 = 2, 2^1+2^2 = 6, 2^1 + 2^2 + 2^3 = 14$ .

In general, the sum of $2^1$ to $2^n = 2^{n+1}-2,$ so the sum of $2^1$ to $2^{100} = 2^{101}-2^1 = 2^1(2^{100} - 1)$ , which is the denominator.

So the $2^{100}-1$ cancels out in the numerator and denominator, leaving us with $\frac{2^{300}}{2^1}$ or $2^{299}$