Problem: 2n+4−2(2n)2(2n+3)\dfrac{2^{n+4}-2\left(2^{n}\right)}{2\left(2^{n+3}\right)}2(2n+3)2n+4−2(2n) when simplified is:
Answer Choices:
A. 2n+1−182^{n+1}-\dfrac{1}{8}2n+1−81
B. −2n+1-2^{n+1}−2n+1
C. 1−2n1-2^{n}1−2n
D. 78\dfrac{7}{8}87
E. 74\dfrac{7}{4}47 Solution:
2n+4−2(2n)2(2n+3)=2n⋅24−2⋅2n2⋅2n⋅23=2⋅2n(23−1)2⋅2n⋅23=78\dfrac{2^{n+4}-2\left(2^{n}\right)}{2\left(2^{n+3}\right)}=\dfrac{2^{n} \cdot 2^{4}-2 \cdot 2^{n}}{2 \cdot 2^{n} \cdot 2^{3}}=\dfrac{2 \cdot 2^{n}\left(2^{3}-1\right)}{2 \cdot 2^{n} \cdot 2^{3}}=\dfrac{7}{8} 2(2n+3)2n+4−2(2n)=2⋅2n⋅232n⋅24−2⋅2n=2⋅2n⋅232⋅2n(23−1)=87