Question

The value of [1^–2 + 2^–2 + 3^–2 ] × 6² is ______.

Answer 1

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 × 6^{2 – 2}

= (7)² × 6⁰  ...[a⁰ – 1]

 = 49