The Value of ${4.1}^{2}$ is equal to

$1.681$

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$168.1$

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$16.81$

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$18.61$

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Solution

The correct option is C
$16.81$

${4.1}^{2} = (\frac{41}{10})^{2}$

$41^{2}$ can be written as $40^{2}$ + (40 +41) [ ( $n + 1)^{2} = n^{2} + (n) + (n + 1)$ ]

So, the expression becomes

$\frac{(40^{2}) + (40 + 41)}{10^{2}}$

= $\frac{1600 + 81}{100}$

= $\frac{1681}{100}$

= $16.81$

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The Value of 4.12 is equal to

The correct option is C 16.81 4.12=(4110)2 412 can be written as 402 + (40 +41) [ (n+1)2=n2+(n)+(n+1) ] So, the expression becomes (402)+(40+41)102 = 1600+81100 = 1681100 = 16.81

The Value of ${4.1}^{2}$ is equal to

The correct option is C
$16.81$

${4.1}^{2} = (\frac{41}{10})^{2}$

$41^{2}$ can be written as $40^{2}$ + (40 +41) [ ( $n + 1)^{2} = n^{2} + (n) + (n + 1)$ ]

So, the expression becomes

$\frac{(40^{2}) + (40 + 41)}{10^{2}}$

= $\frac{1600 + 81}{100}$

= $\frac{1681}{100}$

= $16.81$