[ (22n2^{2n} -3) (22n−22^{2n-2} )( 3n3^{n} -2)( 3n−23^{n-2} )] / 3n−43^{n-4} ( 4n+34^{n+3} - 22n2^{2n} )

asked by guest
on Nov 19, 2024 at 7:14 am



You asked:

Evaluate the expression: (22n−3)⋅22n−2(3n−2)⋅3n−23n−4(4n+3−22n)\frac{\left({2}^{2 n} - 3\right) \cdot {2}^{2 n - 2} \left({3}^{n} - 2\right) \cdot {3}^{n - 2}}{{3}^{n - 4} \left({4}^{n + 3} - {2}^{2 n}\right)}

MathBot Answer:

Evaluated



(22n−3)⋅22n−2(3n−2)⋅3n−23n−4(4n+3−22n)=22n−2⋅34−n3n−2⋅(22n−3)(3n−2)−22n+4n+3\displaystyle \frac{\left({2}^{2 n} - 3\right) \cdot {2}^{2 n - 2} \left({3}^{n} - 2\right) \cdot {3}^{n - 2}}{{3}^{n - 4} \left({4}^{n + 3} - {2}^{2 n}\right)} = \frac{2^{2 n - 2} \cdot 3^{4 - n} 3^{n - 2} \cdot \left(2^{2 n} - 3\right) \left(3^{n} - 2\right)}{- 2^{2 n} + 4^{n + 3}}

Expanded

(22n−3)⋅22n−2(3n−2)⋅3n−23n−4(4n+3−22n)=9⋅24n3n−4⋅22n+256⋅4n−18⋅24n−4⋅22n+256⋅4n−27⋅22n3n−4⋅22n+256⋅4n+54⋅22n−4⋅22n+256⋅4n\frac{\left({2}^{2 n} - 3\right) \cdot {2}^{2 n - 2} \left({3}^{n} - 2\right) \cdot {3}^{n - 2}}{{3}^{n - 4} \left({4}^{n + 3} - {2}^{2 n}\right)} = \frac{9 \cdot 2^{4 n} 3^{n}}{- 4 \cdot 2^{2 n} + 256 \cdot 4^{n}} - \frac{18 \cdot 2^{4 n}}{- 4 \cdot 2^{2 n} + 256 \cdot 4^{n}} - \frac{27 \cdot 2^{2 n} 3^{n}}{- 4 \cdot 2^{2 n} + 256 \cdot 4^{n}} + \frac{54 \cdot 2^{2 n}}{- 4 \cdot 2^{2 n} + 256 \cdot 4^{n}}

Factored

(22n−3)⋅22n−2(3n−2)⋅3n−23n−4(4n+3−22n)=−22n−2⋅34−n3n−2⋅(22n−3)(3n−2)22n−64⋅4n\frac{\left({2}^{2 n} - 3\right) \cdot {2}^{2 n - 2} \left({3}^{n} - 2\right) \cdot {3}^{n - 2}}{{3}^{n - 4} \left({4}^{n + 3} - {2}^{2 n}\right)} = - \frac{2^{2 n - 2} \cdot 3^{4 - n} 3^{n - 2} \cdot \left(2^{2 n} - 3\right) \left(3^{n} - 2\right)}{2^{2 n} - 64 \cdot 4^{n}}

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