Find the value of 32 - 4-1 - 22

Given , (3^2 + 4^ 1) × 2^2, ⇒ (9 + 1/4) × 2^2 (∵ a^ m = 1/a^m) = 36 + 1/4 × 4 = 37/4 × 4 = 37 ...

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Byju's Answer

Standard IX

Mathematics

Conversion of Numbers Using Logarithms

Find the valu...

Question

Find the value of ( $3^{2}$ + $4^{- 1}$ ) $\times$ $2^{2}$

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Solution

The correct option is A
37

Given , ( $3^{2}$ + $4^{- 1}$ ) $\times$ $2^{2}$ ,

$\Rightarrow$ $(9 + \frac{1}{4}$ ) $\times$ $2^{2}$ ( $∵ a^{- m} = \frac{1}{a^{m}}$ )

= $\frac{36 + 1}{4}$ $\times$ 4

= $\frac{37}{4}$ $\times$ 4

= 37

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Similar questions

Q. Observe the following pattern

1^{2} = \frac{1}{6} [1 \times (1 + 1) \times (2 \times 1 + 1)] 1^{2} + 2^{2} = \frac{1}{6} [2 \times (2 + 1) \times (2 \times 2 + 1)] 1^{2} + 2^{2} + 3^{2} = \frac{1}{6} [3 \times (3 + 1) \times (2 \times 3 + 1)] 1^{2} + 2^{2} + 3^{2} + 4^{2} = \frac{1}{6} [4 \times (4 + 1) \times (2 \times 4 + 1)]

and find the values of each of the following:
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$1 + \frac{1^{2} + 2^{2}}{2!} + \frac{1^{2} + 2^{2} + 3^{2}}{3!} + \frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{4!} + . . . \infty = \frac{b}{6} e$ . Find $b$

Q. The value of the infinite series

\frac{1^{2} + 2^{2}}{3!} + \frac{1^{2} + 2^{2} + 3^{2}}{4!} + \frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{5!} + . . . . . .

Q. If,

\sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + \sqrt{1 + \frac{1}{3^{2}} + \frac{1}{4^{2}}} + . . . . . + \sqrt{1 + \frac{1}{(1999)^{2}} + \frac{1}{(2000)^{2}}} = x - \frac{1}{x}

, then find the value of

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Find the value of (32 + 4−1) × 22

The correct option is A 37 Given , (32 + 4−1) × 22, ⇒ (9 + 14) × 22 (∵a−m = 1am) = 36 + 14 × 4 = 374 × 4 = 37

Find the value of ( $3^{2}$ + $4^{- 1}$ ) $\times$ $2^{2}$

The correct option is A
37

Given , ( $3^{2}$ + $4^{- 1}$ ) $\times$ $2^{2}$ ,

$\Rightarrow$ $(9 + \frac{1}{4}$ ) $\times$ $2^{2}$ ( $∵ a^{- m} = \frac{1}{a^{m}}$ )

= $\frac{36 + 1}{4}$ $\times$ 4

= $\frac{37}{4}$ $\times$ 4

= 37