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Algebraic Identities

Question

Evaluate $(41)^{2}$ by using the identity $(a + b)^{2} = a^{2} + 2 a b + b^{2}$

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Solution

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Using $(a + b)^{2}$ = $a^{2} + 2 a b + b^{2}$
Writing $(41)^{2}$ as $(40 + 1)^{2}$
$=$ $40^{2} + 2 \times 40 + 1 + 1^{2}$
$=$ $1600 + 80 + 1$
The solution is $1681$

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