0.2+0.22+0.022+ . up to n terms is equal to

Correct Answer is : 29[n−19(1−10−n)]2/9≤ft[n-1/9≤ft(1-10^{-n})]92[n−91(1−10−n)]

2/9≤ft[n-1/9≤ft(1-10^{-n})]

Correct Answer is : 29[n−19(1−10−n)]2/9≤ft[n-1/9≤ft(1-10^{-n})]92[n−91(1−10−n)]

Test Index

COMEDK 2025 Morning Solved Paper

Question : 124 of 180

Marks: +1, -0

0.2+0.22+0.022+\ldots \ldots \ldots

n

terms is equal to

Solution:

Concept:

Each term is

2 \times 0.\underbrace{111\ldots1}_{k\text{ times}}

which equals

\frac{2}{9}(1-10^{-k})

Explanation:

The series is

0.2 + 0.22 + 0.222 + \dots

up to

n

terms.

The

k

-th term is

a_k = 0.\underbrace{22\ldots2}_{k\text{ times}} = 2 \times 0.\underbrace{11\ldots1}_{k\text{ times}}

Now

0.\underbrace{11\ldots1}_{k\text{ times}} = \frac{10^k - 1}{9 \cdot 10^k} = \frac{1-10^{-k}}{9}

Thus

a_k = \frac{2}{9} (1-10^{-k})

Sum up to

n

terms:

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

The geometric series

\sum_{k=1}^n 10^{-k} = \frac{10^{-1}(1-10^{-n})}{1-10^{-1}} = \frac{1-10^{-n}}{9}

Substituting gives

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

Answer:

Option B:

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

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