Question

show that
$\frac { 16 \times 2 ^ { n - 1 } - 4 \times 2 ^ { n } } { 16 \times 2 ^ { n + 2 } - 2 \times 2 ^ { n + 2 } } = \frac { 1 } { 14 }$
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Answer 1

Step-by-step explanation:

show that

\frac { 16 \times 2 ^ { n - 1 } - 4 \times 2 ^ { n } } { 16 \times 2 ^ { n + 2 } - 2 \times 2 ^ { n + 2 } } = \frac { 1 } { 14 }

16×2

n+2

−2×2

n+2

16×2

n−1

−4×2

n

=

14

1

please don't spam I have interest that what's your opinion about the question

Answer 2

$\sf\small\underline\green{Given:-}$

$\rm{\implies \dfrac{16\times\:2^{n-1}-4\times\:2^{n}}{16\times\:2^{n+2}-2\times\:2^{n+2}}=\dfrac{1}{14}}$

$\sf\small\underline\green{Solution:-}$

$\rm{\implies \dfrac{16\times\:2^{n-1}-4\times\:2^{n}}{16\times\:2^{n+2}-2\times\:2^{n+2}}=\dfrac{1}{14}}$

$\sf{\implies \dfrac{16\times\:2^n-2^{-1}-4\times\:2^n}{16\times\:2^n*2^2-2\times\:2^n*2^2}=\dfrac{1}{14}}$

$\sf{\implies \dfrac{2^n(16\times\:2^{-1}-4)}{2^n(16\times\:2^2-2\times\:2^2)}=\dfrac{1}{14}}$

$\sf{\implies \dfrac{2^3-4}{16*4-2*4}}$

$\sf{\implies \dfrac{4}{64-8}}$

$\sf{\implies \dfrac{4}{56}}$

$\sf{\implies \dfrac{1}{14}}$