Question on: WAEC Mathematics - 2014
Simplify \(\frac{8^2 \times 4^n + 1}{2^{2n} \times 16}\)
A
16
B
8
C
4
D
1
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Correct Option: C
\(\frac{\sqrt{8^2 \times 4^{n + 1}}}{2^{2n} \times 16}\)
= \(\frac{\sqrt{2^{3(2)} \times 2^{2(n + 1)}}}{2^{2n} \times 2^4}\)
= \(\frac{\sqrt{2^6 \times 2^{2n + 2)}}}{2^{2n} + 4}\)
= \(\frac{\sqrt{2^6 + 2^{2n + 2)}}}{2^{2n} + 4}\)
= \(\frac{\sqrt{2^{2n + 8}}}{2^{2n} + 4}\)
= \(\sqrt{2^{2n + 8} \div 2^{2n} + 4}\)
= \(\sqrt{2^{2n - 2n} + 8 - 4}\)
= \(\sqrt{2^4}\)
= \(\sqrt{16}\)
= 4
= \(\frac{\sqrt{2^{3(2)} \times 2^{2(n + 1)}}}{2^{2n} \times 2^4}\)
= \(\frac{\sqrt{2^6 \times 2^{2n + 2)}}}{2^{2n} + 4}\)
= \(\frac{\sqrt{2^6 + 2^{2n + 2)}}}{2^{2n} + 4}\)
= \(\frac{\sqrt{2^{2n + 8}}}{2^{2n} + 4}\)
= \(\sqrt{2^{2n + 8} \div 2^{2n} + 4}\)
= \(\sqrt{2^{2n - 2n} + 8 - 4}\)
= \(\sqrt{2^4}\)
= \(\sqrt{16}\)
= 4
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