Find the square of $(101)^{2}$

10, 102

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10, 202

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10, 101

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10, 201

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Solution

The correct option is D
10, 201

$(101)^{2} = (100)^{2} + 100 + 101 [(n + 1)^{2} = n^{2} + (n) + (n + 1)]$

$(101)^{2} = 10, 000 + 201$

$(101)^{2} = 10201$

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Similar questions

(101)^{2}

How many non square numbers lie between the following pairs of the numbers.

1. 100²and 101²

101^{2} - 99^{2} =

m

(- 10)^{m - 3} \times (- 10)^{5} = (- 10)^{12}

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