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Question
The value of [1–2 + 2–2 + 3–2 ] × 62 is ______.
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Solution
The value of [1–2 + 2–2 + 3–2 ] × 62 is 49.
Explanation:
Using law of exponents,
`a^-m = 1/a^m` ...[∵ a is non-zero integer]
∴ `[1^-2 + 2^-2 + 3^-2] xx 6^2 = [1/1^2 + 1/2^2 + 1/3^2] xx 6^2`
= `[1 + 1/4 + 1/9] xx 6^2`
= `((36 + 9 + 4)/36) xx 6^2`
= `(49/36) xx 6^2`
= `(7/6)^2 xx 6^2`
= (7)2 × 6–2 × 62
= (7)2 × 62 – 2
= (7)2 × 60 ...[a0 – 1]
= 49
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