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

Kerala DhseKerala DHSE Plus One Board 2022Subjective· 4mImportance★★★★★
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Factor out 8, rewrite each repunit-like term 11…1⏟k\underbrace{11\ldots1}_{k} as 10k−19\dfrac{10^k-1}{9}, then sum the resulting GP.

Sn=8+88+888+⋯S_n = 8+88+888+\cdots to nn terms.

Factor 88 from every term:

Sn=8(1+11+111+⋯ to n terms)S_n = 8(1+11+111+\cdots\text{ to }n\text{ terms})

Each term 11…1⏟k ones=10k−19\underbrace{11\ldots1}_{k\text{ ones}} = \dfrac{10^k-1}{9}. So:

1+11+111+⋯(n terms)=19[(10−1)+(102−1)+⋯+(10n−1)]=19[(10+102+⋯+10n)−n]1+11+111+\cdots(n\text{ terms}) = \dfrac19\left[(10-1)+(10^2-1)+\cdots+(10^n-1)\right] = \dfrac19\left[(10+10^2+\cdots+10^n)-n\right]

The sum 10+102+⋯+10n10+10^2+\cdots+10^n is a GP (first term 1010, ratio 1010, nn terms):

10+102+⋯+10n=10(10n−1)10−1=10n+1−10910+10^2+\cdots+10^n = \dfrac{10(10^n-1)}{10-1} = \dfrac{10^{n+1}-10}{9}

So: …

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