Use identities to evaluate: $(101)^{2}$

11, 601

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12, 761

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10, 111

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10, 201

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Solution

The correct option is D $10, 201$
$(101)^{2}$
$= (100 + 1)^{2}$
using, $(a + b)^{2} = a^{2} + 2 a b + b^{2}$
$= (100)^{2} + 2 (100) (1) + 1^{2}$
$= 10000 + 200 + 1$
$= 10201$

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