Question

Question 65

In the following question, fill in the blanks to make the statements true.

The value of $[1^{-2}+2^{-2}+3^{-2}]\times 6^2$ is ___

Answer 1

Using law of exponents,$a^{-m}=\frac{1}{a^m}\therefore [1^{-2}+2^{-2}+3^{-2}]\times 6^2=[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}]\times 6^2=[1+\frac{1}{4}+\frac{1}{9}]\times 6^2=\left(\frac{36+9+4}{36}\right)\times 6^2=\left(\frac{49}{36}\right)\times 6^2={\left(\frac{7}{6}\right)}^2\times 6^2=(7{)}^2\times 6^{-2}\times 6^2=(7{)}^2\times 6^{2-2}=(7{)}^2\times 6^0=49$
Hence$[1^{-1}+2^{-2}+3^{-2}]\times 6^2=49$