Using the identity nbsp;(a + b)^2= nbsp;a^2 +2ab + b^2, we have (101)^2 nbsp;=(100+1)^2 nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nb ...

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Algebraic Identities

Evaluate :101...

Question

Evaluate : $101^{2}$

10,209

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10,201

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10,401

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10,211

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Solution

The correct option is B
10,201

Using the identity $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ , we have
$(101)^{2}$ $= (100 + 1)^{2}$
$= (100)^{2} + 2 (100) (1) + (1)^{2}$
$= 10000 + 200 + 1$
$= 10, 201$

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