The value of -1-2 - -1-2-2 -1-3-2- -1-4-2-is-28-25-60-30-

The correct option is D 30 (1)^ 2 + (1/2)^ 2 +(1/3)^ 2+ (1/4)^ 2 (1)^ 2 + (2^ 1)^ 2 +(3^ 1)^ 2+ (4^ 1)^ 2 (∵ a^ m = 1/a^m) = (1)^ 2 + (2)^( 1 × 2) +(3)^( ...

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Byju's Answer

Standard IX

Mathematics

Exponent

The value of ...

Question

The value of $(1)^{- 2}$ + ${(\frac{1}{2})}^{- 2}$ + ${(\frac{1}{3})}^{- 2}$ + ${(\frac{1}{4})}^{- 2}$ is:

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Solution

The correct option is C
30

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

$(1)^{- 2}$ + $(2^{- 1})^{- 2}$ + $(3^{- 1})^{- 2}$ + $(4^{- 1})^{- 2}$ ( $∵ a^{- m} = \frac{1}{a^{m}}$ )

= $(1)^{- 2}$ + $(2)^{(- 1 \times - 2)}$ + $(3)^{(- 1 \times - 2)}$ + $(4)^{(- 1 \times - 2)}$ ( $∵ (a^{n})^{m} = (a)^{m \times n}$ )
= 1 + $2^{2}$ + $3^{2}$ + $4^{2}$
= 1 + 4 + 9 + 16
= 30

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The value of (1)−2 + (12)−2 +(13)−2+ (14)−2 is:

The correct option is C 30 (1)−2 + (12)−2 +(13)−2+ (14)−2 (1)−2 + (2−1)−2 +(3−1)−2+ (4−1)−2 (∵a−m = 1am) = (1)−2 + (2)(−1 × −2) +(3)(−1 × −2)+ (4)(−1 × −2) (∵(an)m = (a)m×n) = 1 + 22+ 32 + 42 = 1 + 4 + 9 + 16 = 30

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