Question 73

Find the value of $n$ , where $n$ is an integer and $2^{n - 5} \times 6^{2 n - 4} = \frac{1}{12^{4} \times 2} .$

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Solution

We have $2^{n - 5} \times 6^{2 n - 4} = \frac{1}{12^{4} \times 2} \Rightarrow \frac{2^{n}}{2^{5}} \times \frac{6^{2 n}}{6^{4}} = \frac{1}{12^{4} \times 2} \Rightarrow \frac{2^{n} \times 6^{2 n}}{2^{5} \times 6^{4}} = \frac{1}{(2 \times 6)^{4} \times 2} \Rightarrow 2^{n} \times (6^{2})^{n} = \frac{2^{5} \times 6^{4}}{2^{4} \times 6^{4} \times 2} \Rightarrow 2^{n} \times 36^{n} = \frac{2^{5} \times 6^{4}}{2^{5} \times 6^{4}} \Rightarrow 2^{n} \times 36^{n} = 1 \Rightarrow (2 \times 36)^{n} = 1 \Rightarrow (72)^{n} = (72)^{0} ∴ n = 0$

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2^{n - 5} \times 6^{2 n - 4} = \frac{1}{12^{4} \times 2}

Find the value of n, Where n is an integer and 2^n-2 multiplied6^2n-4=1/(12^4 multiplied2)

2^{n} - (n - 1) 2^{n - 2} + \frac{(n - 2) (n - 3)}{⌊ 2} - \frac{(n - 3) (n - 4) (n - 5)}{⌊ 3} 2^{n - 6} + . . . . .;

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a = \sum_{n = 1}^{\infty} (\frac{2 n}{(2 n - 1)!})

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