Powers and Roots – Simplify an expression with powers – Exercise 1653

Exercise

Simplify the expression:

2−n+1⋅4n−1+2n2n+3−2n−1\frac{2^{-n+1}\cdot 4^{n-1}+2^n}{2^{n+3}-2^{n-1}}

Final Answer

2−n+1⋅4n−1+2n2n+3−2n−1=15\frac{2^{-n+1}\cdot 4^{n-1}+2^n}{2^{n+3}-2^{n-1}}=\frac{1}{5}

Solution

Using Powers and Roots rules we get:

2−n+1⋅4n−1+2n2n+3−2n−1=\frac{2^{-n+1}\cdot 4^{n-1}+2^n}{2^{n+3}-2^{n-1}}=

=2−n+1⋅(22)n−1+2n2n⋅23−2n⋅2−1==\frac{2^{-n+1}\cdot {(2^2)}^{n-1}+2^n}{2^n\cdot 2^3-2^n\cdot 2^{-1}}=

=2−n+1⋅22n−2+2n2n⋅23−2n⋅2−1==\frac{2^{-n+1}\cdot 2^{2n-2}+2^n}{2^n\cdot 2^3-2^n\cdot 2^{-1}}=

=2−n+1+2n−2+2n2n(23−2−1)==\frac{2^{-n+1+2n-2}+2^n}{2^n(2^3-2^{-1})}=

=2n⋅2−1+2n2n(23−2−1)==\frac{2^n\cdot 2^{-1}+2^n}{2^n(2^3-2^{-1})}=

=2n(2−1+1)2n(23−2−1)==\frac{2^n(2^{-1}+1)}{2^n(2^3-2^{-1})}=

=2−1+123−2−1==\frac{2^{-1}+1}{2^3-2^{-1}}=

=12+18−12==\frac{\frac{1}{2}+1}{8-\frac{1}{2}}=

=32152==\frac{\frac{3}{2}}{\frac{15}{2}}=

=32⋅215==\frac{3}{2}\cdot\frac{2}{15}=

=630=15=\frac{6}{30}=\frac{1}{5}

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