0.2+0.22+0.222+ to n terms is equal to

Correct Answer is : (29)≤ft(2/9)(92)

≤ft(2/9)≤ft[n-≤ft(1/9)≤ft(1-10^{-n})]

Correct Answer is : (29)≤ft(2/9)(92)

Test Index

Manipal Entrance Test 2017 Math Paper

Question : 46 of 70

Marks: +1, -0

0.2+0.22+0.222+\dots

n

terms is equal to

$n-\left(\frac{1}{9}\right)\left(1-10^{-n}\right)$
$\left(\frac{2}{9}\right)\left[n-\left(\frac{1}{9}\right)\left(1-10^{-n}\right)\right]$
$\left(\frac{2}{9}\right)$
$(2/9)$

Solution:

0.2 + 0.22 + 0.222 + ... n terms

= 2(0.1 + 0.11 + 0.111 + .. . n term

=2\left(\frac{1}{10}+ \frac{11}{100}+ \frac{111}{1000}+\dots \text{ n terms }\right)

=\frac{2}{9}\left(\frac{9}{10}+ \frac{99}{100}+ \frac{999}{1000}+\dots \text{ n terms }\right)

=\frac{2}{9}\left(1-\frac{1}{10}+1-\frac{1}{100}+1-\frac{1}{1000}+\dots\text{ n terms}\right)

=\frac{2}{9}\left[n-\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\dots n\right)\right]

=\frac{2}{9} \left[ n - \frac{1}{10} \cdot \frac{ \left\{ 1- \left(\frac{1}{10}\right)^{n} \right\} }{ \left(1- \frac{1}{10}\right) } \right]

=\frac{2}{9}\left[ n - \frac{1}{10} \times \frac{10}{9} \cdot \left( \frac{10^{n}-1}{10^{n}} \right) \right]

=\left(\frac{2}{9}\right)\left[n-\left(\frac{1}{9}\right)\left(1-10^{-n}\right)\right]

\left(\frac{2}{9}\right)-\left(\frac{2}{81}\right)\left(1-10^{-n}\right)

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