$\frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} + 2^{- 3}$ , is equal to

2^{n + 1}

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- 2^{n + 1} + \frac{1}{8}

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\frac{9}{8} - 2^{n}

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1

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Solution

The correct option is D $1$
$\frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} + 2^{- 3}$

$\frac{2^{n} (2^{4} - 2)}{2^{n} {.2}^{4}} + \frac{1}{2^{3}}$

$\frac{16 - 2}{16} + \frac{1}{8}$

$\frac{14 + 2}{16} = \frac{16}{16} = 1$

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Similar questions

\frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} + 2^{- 3}

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

\sum \sum 0 \leq r < s \leq n (r + s) C_{r} C_{s}

\frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} =

{lim}_{n \to \infty} (\frac{n}{(n + 1) \sqrt{2 n + 1}} + \frac{n}{(n + 2) \sqrt{2 (2 n + 2)}} + \frac{n}{(n + 3) \sqrt{3 (2 n + 3)}} . . . . . + \frac{1}{2 n \sqrt{3}})

\frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})}

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