Question

$\frac{2^{n+4}-{2.2}^n}{{2.2}^{n+3}}+2^{-3}$ is equal to

• A. 2 ⁿ⁺¹
• B. -2ⁿ⁺¹ + 1/8
• C. 9/8(2ⁿ)
• D. 1

Answer 1

The correct option is D

1

The~given~expression

$\begin{array}{c}\frac{2^n.2^4-2.2^n}{2.2^n.2^3}+\frac{1}{2^3}\\ =\frac{2^n.\left(2^4-2\right)}{2.\left(2^n.2^3\right)}+\frac{1}{2^3}=\left(\frac{16-2}{16}+\frac{1}{8}\right)\\ =\left(\frac{7}{8}+\frac{1}{8}\right)=1.\end{array}$