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Exercise 8.2 · Q18

Q.Find the sum to nn terms of the sequence, 8,88,888,8888,…8, 88, 888, 8888, \ldots

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The sequence 8,88,888,…8, 88, 888, \dots is not a standard GP, but each term can be written as 89(10k−1)\frac{8}{9}(10^k - 1). Summing these gives Sn=881(10n+1−9n−10)S_n = \frac{8}{81}\left(10^{n+1} - 9n - 10\right).

This problem looks like a geometric progression at first glance — the terms grow rapidly, and each term seems to be built by appending an 8. But check the ratio: 88/8=1188/8 = 11, 888/88=10.09…888/88 = 10.09\ldots, not constant. So it’s not a GP. The trick is to see the pattern in terms of powers of 10.

Each term is a string of 8’s. For example:

  • 8=8×18 = 8 \times 1
  • 88=8×1188 = 8 \times 11
  • 888=8×111888 = 8 \times 111
  • 8888=8×11118888 = 8 \times 1111

And 111…111\ldots (k times) can be written as 10k−19\frac{10^k - 1}{9}. So the kk-th term is Tk=89(10k−1)T_k = \frac{8}{9}(10^k - 1).

Now the sum to nn terms becomes a sum of two parts: a geometric series in 10k10^k and a constant term.

  1. Write the general term

    Tk=89(10k−1)T_k = \frac{8}{9}(10^k - 1) for k=1,2,…,nk = 1, 2, \dots, n.

  2. Sum over k

    Sn=∑k=1nTk=89(∑k=1n10k−∑k=1n1)S_n = \sum_{k=1}^n T_k = \frac{8}{9} \left( \sum_{k=1}^n 10^k - \sum_{k=1}^n 1 \right).

  3. Evaluate the geometric sum

    ∑k=1n10k=10+102+⋯+10n=10(10n−1)10−1=10(10n−1)9\sum_{k=1}^n 10^k = 10 + 10^2 + \dots + 10^n = \frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}.

  4. Evaluate the constant sum

    ∑k=1n1=n\sum_{k=1}^n 1 = n.

  5. Combine

    Sn=89(10(10n−1)9−n)=89⋅10(10n−1)−9n9S_n = \frac{8}{9} \left( \frac{10(10^n - 1)}{9} - n \right) = \frac{8}{9} \cdot \frac{10(10^n - 1) - 9n}{9}.

    So Sn=881(10n+1−10−9n)S_n = \frac{8}{81} \left( 10^{n+1} - 10 - 9n \right). …

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