Evaluate $(101)^{2}$ using identity $(a+b)^{2}=a^{2}+2ab+b^{2}$ with $a=100,\ b=1$.

Evaluate $(101)^{2}$ using identity $(a+b)^{2}=a^{2}+2ab+b^{2}$ with $a=100,\ b=1$.
  1. $10201$
  2. $10001$
  3. $10101$
  4. $11011$

Solution

$(101)^{2} = (100+1)^{2} = 10000 + 200 + 1 = 10201$.

Asked in: IMO

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