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Miscellaneous Exercise · Q11

Q.Find the sum of the following series up to nn terms:

(i) 5+55+555+…5 + 55 + 555 + \ldots
(ii) 0.6+0.66+0.666+…0.6 + 0.66 + 0.666 + \ldots
Andaman Nicobar CbseNCERTSubjective· 3mImportance★★★★★est
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Both series are built from repeating digits. The key is to rewrite each term as a multiple of a geometric series: 5+55+555+⋯=59(101−1+102−1+… )5 + 55 + 555 + \dots = \frac{5}{9}(10^1 - 1 + 10^2 - 1 + \dots) and 0.6+0.66+0.666+⋯=69(1−10−1+1−10−2+… )0.6 + 0.66 + 0.666 + \dots = \frac{6}{9}(1 - 10^{-1} + 1 - 10^{-2} + \dots). Summing the geometric parts gives closed forms: (i) 581(10n+1−9n−10)\frac{5}{81}(10^{n+1} - 9n - 10) and (ii) 227(9n−1+10−n)\frac{2}{27}(9n - 1 + 10^{-n}).


Why this approach works

When you see a series like 5,55,555,…5, 55, 555, \dots, the pattern is obvious — each term appends another digit 5. But adding these directly is messy. The insight is to notice that 55=5×1155 = 5 \times 11, 555=5×111555 = 5 \times 111, and so on. And 111…111\ldots (with kk ones) is exactly 10k−19\frac{10^k - 1}{9}. That turns the problem into a sum of powers of 10, which is a geometric series — something we know how to handle cleanly.

The second series 0.6,0.66,0.666,…0.6, 0.66, 0.666, \dots is the same idea, just shifted to the right of the decimal point. 0.666…0.666\ldots (with kk sixes) equals 69(1−10−k)\frac{6}{9}(1 - 10^{-k}). Again, the sum becomes a geometric series in 10−110^{-1}.


(i) 5+55+555+…5 + 55 + 555 + \ldots up to nn terms

Step 1: Write each term using the "repunit" trick

A number with kk digits all equal to 5 can be written as:

5×11…1⏟k ones=5×10k−195 \times \underbrace{11\ldots1}_{k \text{ ones}} = 5 \times \frac{10^k - 1}{9}

So the kk-th term of the series is:

Tk=59(10k−1)T_k = \frac{5}{9}(10^k - 1)

Step 2: Sum from k=1k=1 to nn

Let SnS_n be the sum of the first nn terms:

Sn=∑k=1nTk=59∑k=1n(10k−1)S_n = \sum_{k=1}^n T_k = \frac{5}{9} \sum_{k=1}^n (10^k - 1)

Separate the sum:

Sn=59(∑k=1n10k−∑k=1n1)S_n = \frac{5}{9} \left( \sum_{k=1}^n 10^k - \sum_{k=1}^n 1 \right)

Step 3: Evaluate each part

The first sum is a geometric series with first term 1010, common ratio 1010:

∑k=1n10k=10⋅10n−110−1=10(10n−1)9\sum_{k=1}^n 10^k = 10 \cdot \frac{10^n - 1}{10 - 1} = \frac{10(10^n - 1)}{9}

The second sum is just nn ones:

∑k=1n1=n\sum_{k=1}^n 1 = n

Step 4: Combine

Sn=59(10(10n−1)9−n)S_n = \frac{5}{9} \left( \frac{10(10^n - 1)}{9} - n \right)

Put over a common denominator 99:

Sn=59⋅10(10n−1)−9n9=581(10n+1−10−9n)S_n = \frac{5}{9} \cdot \frac{10(10^n - 1) - 9n}{9} = \frac{5}{81} \left( 10^{n+1} - 10 - 9n \right)

Simplify:

Sn=581(10n+1−9n−10)S_n = \frac{5}{81} \left( 10^{n+1} - 9n - 10 \right)

Tip

A quick check: for n=1n=1, S1=5S_1 = 5. Plug n=1n=1 into the formula: 581(102−9−10)=581(100−19)=581×81=5\frac{5}{81}(10^2 - 9 - 10) = \frac{5}{81}(100 - 19) = \frac{5}{81} \times 81 = 5. Works.


(ii) 0.6+0.66+0.666+…0.6 + 0.66 + 0.666 + \ldots up to nn terms

Step 1: Rewrite each term

A decimal like 0.666…0.666\ldots with kk sixes after the decimal point is:

0.66…6⏟k digits=69(1−10−k)0.\underbrace{66\ldots6}_{k \text{ digits}} = \frac{6}{9}(1 - 10^{-k})

Why? Because 0.6‾=690.\overline{6} = \frac{6}{9}, and 0.666…0.666\ldots with kk sixes is 69\frac{6}{9} minus the tail that starts after the kk-th place. That tail is 69×10−k\frac{6}{9} \times 10^{-k}.

So the kk-th term is:

Tk=69(1−10−k)=23(1−10−k)T_k = \frac{6}{9}(1 - 10^{-k}) = \frac{2}{3}(1 - 10^{-k}) …

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