The value of $(2^{3} + 2^{2} + 2^{- 2} + 2^{- 3})$ is equal to _____.

\frac{99}{8}

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\frac{99}{16}

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\frac{97}{8}

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6

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Solution

The correct option is B $\frac{99}{8}$
$2^{3} + 2^{2} + 2^{- 2} + 2^{- 3}$

$\Rightarrow$ $8 + 4 + \frac{1}{2^{2}} + \frac{1}{2^{3}}$

$\Rightarrow$ $12 + \frac{1}{4} + \frac{1}{8}$

$\Rightarrow$ $\frac{96 + 2 + 1}{8}$

$\Rightarrow$ $\frac{99}{8}$

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Similar questions

Q. The value of

100^{2} - 99^{2} + 98^{2} - 97^{2} . . . . . . . . . - 3^{2} + 2^{2} - 1^{2}

is:

State True or False.

(125)^{- \frac{2}{3}} \div (8)^{- \frac{2}{3}} = \frac{1}{99}

The remainder obtained when $(2 + 5^{2} + 6^{3} + 11^{4} + 2^{5} + 5^{6} + 6^{7} + 11^{8} + \dots \dots 2^{97} + 5^{98} + 6^{99} + 11^{100})$ is divided by 5 is

Q. Multiplication Patterns

9 × 1 = 9
9 × 2 = 18	1 + 8 = 9
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Did you see the pattern in the 9 times table? What numbers are adding up to 9?

Q. What is the remainder when p - q is divided by 100, if

p = 2^{99} + 4^{99} + 6^{99} + 8^{99} + . . . . . . . . . + 100^{99}

and

q = 1^{99} + 3^{99} + 5^{99} + 7^{99} + . . . . . . . . . . + 99^{99} ?

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The value of (23+22+2−2+2−3) is equal to _____.

The correct option is B 99823+22+2−2+2−3⇒ 8+4+122+123⇒ 12+14+18⇒ 96+2+18⇒ 998

The value of $(2^{3} + 2^{2} + 2^{- 2} + 2^{- 3})$ is equal to _____.

The correct option is B $\frac{99}{8}$
$2^{3} + 2^{2} + 2^{- 2} + 2^{- 3}$

$\Rightarrow$ $8 + 4 + \frac{1}{2^{2}} + \frac{1}{2^{3}}$

$\Rightarrow$ $12 + \frac{1}{4} + \frac{1}{8}$

$\Rightarrow$ $\frac{96 + 2 + 1}{8}$

$\Rightarrow$ $\frac{99}{8}$