Using standard identity, find the value of $(a + 2 b)^{2}$ .

a^{2} + 2 b^{2} + 4 a b

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a^{2} + 4 b^{2} + 2 a b

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a^{2} + 4 b^{2} + 4 a b

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2 a^{2} + 4 b^{2} + 4 a b

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Solution

The correct option is D $a^{2} + 4 b^{2} + 4 a b$
Given, $(a + 2 b)^{2}$ .

We know, $(x + y)^{2} = x^{2} + 2 x y + y^{2}$ .

Then,
$(a + 2 b)^{2}$ $= a^{2} + (2 b)^{2} + 2 \times a \times 2 b$
$= a^{2} + 4 b^{2} + 4 a b$ .

Therefore, option $C$ is correct.

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