Find the value of $(201)^{2}$

$40, 104$

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$40, 401$

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$40, 441$

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$40, 414$

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Solution

The correct option is B
$40, 401$

$(201)^{2} = (200)^{2} + 200 + 201 [(n + 1)^{2} = n^{2} + (n) + (n + 1)]$

$(201)^{2} = 40, 000 + 401$

$(201)^{2} = 40, 401$

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Find the value of (201)2

The correct option is B 40,401 (201)2=(200)2+200+201 [(n+1)2=n2+(n)+(n+1)] (201)2=40,000+401 (201)2=40,401

Find the value of $(201)^{2}$

The correct option is B
$40, 401$

$(201)^{2} = (200)^{2} + 200 + 201 [(n + 1)^{2} = n^{2} + (n) + (n + 1)]$

$(201)^{2} = 40, 000 + 401$

$(201)^{2} = 40, 401$