Question

Using suitable identities evaluate: ${99}^2$

Answer 1

${99}^2={(100-1)}^2$ using ${(a-b)}^2=a^2+b^2-2ab$

where $a=100,b=1$

$={100}^2+1^2-2\times 100\times 1$

$=10000+1-200=9801‬$