Question

`((2^(2n)-3*2^(2n-2))(3^(n)-2*3^(n-2)))/(3^(n-4)(4^(n+3)-2^(2n)))​

Answer 1

Answer:

$\frac{\left(2^{2n}-3\cdot \:2^{2n-2}\right)\left(3^n-2\cdot \:3^{n-2}\right)}{3^{n-4}\left(4^{n+3}-2^{2n}\right)}\\= \frac{\left(1-\frac{3}{4}\right)\left(1-\frac{2}{9}\right)\cdot \:12^n}{3^{n-4}\left(4^{n+3}-2^{2n}\right)}\\= \frac{\left(-\frac{3}{4}+1\right)\left(-\frac{2}{9}+1\right)\cdot \:3^n\cdot \:2^{2n}}{3^{n-4}\left(4^{n+3}-2^{2n}\right)}\\=\frac{3^4\cdot \frac{1}{4}\left(-\frac{2}{9}+1\right)\cdot \:2^{2n}}{4^{n+3}-2^{2n}}$

$= \frac{3^4\cdot \frac{1}{4}\cdot \frac{7}{9}\cdot \:2^{2n}}{4^{n+3}-2^{2n}}\\= \frac{81\cdot \frac{1}{4}\cdot \frac{7}{9}\cdot \:2^{2n}}{4^{n+3}-2^{2n}}\\= \frac{63\cdot \:2^{2n-2}}{63\cdot \:4^n}\\= \frac{2^{2n-2}}{2^{2n}}\\= \frac{1}{4}$

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