If $1^{2} + 2^{2} + . . . + 9^{2} = 285,$ then the value of $(0.11)^{2} + (0.22)^{2} + . . . . + (0.99)^{2}$ is

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Solution

${(0.11)}^{2} + {(0.22)}^{2} + . . . . . . {(0.99)}^{2}$
$= {(\frac{11}{100} \times 1)}^{2} + {(\frac{11}{100} \times 2)}^{2} + . . . . . . . . . . {(\frac{11}{100} \times 9)}^{2}$
$= {(\frac{11}{100})}^{2} [1^{2} + 2^{2} + 3^{2} + . . . . . . . . 9^{2}]$
$= \frac{121}{10000} \times 285 = 3.4485$

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