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Question 67
David is trying to solve the expression :-
$$\frac{(4)^2 \times 2^{n + 1} - 4 \times 2^n}{(4)^2 \times 2^{n + 2} - 2 \times 2^{n + 2}}$$
And you help him to do the same and finally arrive at the answer with correct to one decimal which would be -
(Note:- DO NOT include spaces in your answer)
Correct Answer: 0.5
Solution
$$\frac{(4)^2 \times 2^{n + 1} - 4 \times 2^n}{(4)^2 \times 2^{n + 2} - 2 \times 2^{n + 2}}$$
$$\frac{(2)^4 \times 2^{n + 1} - 2^2 \times 2^n}{(2)^4 \times 2^{n + 2} - 2 \times 2^{n + 2}}$$
$$\frac{(2)^3 \times 2^{n}*2 -2 \times 2^n}{(2)^3 \times 2^{n}*2^2 - 2^{n}*2^2}$$
$$\ \frac{\ 2^3\times\ 2-2}{2^3\cdot2^2-2^2}$$
$$\ \frac{\ 14}{28}$$
=0.5
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