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Q.Find the sum to nn terms of the sequence 7,77,777,7777,…7, 77, 777, 7777, \ldots to nn terms.

Himachal HpboseHPBOSE Himachal Pradesh Class 11 Board Exam 2024Subjective· 5mImportance★★★★★
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The sum to nn terms of 7,77,777,…7,77,777,\ldots is Sn=781(10n+1−9n−10)S_n=\dfrac{7}{81}\left(10^{n+1}-9n-10\right).

The kkth term is a kk-digit number of all 77s: 7, 77, 777,…7,\ 77,\ 777,\ldots. Factor out 77 from every term:

Sn=7+77+777+⋯(n terms)=7 (1+11+111+⋯(n terms)).S_n=7+77+777+\cdots\text{(}n\text{ terms)}=7\,(1+11+111+\cdots\text{(}n\text{ terms)}).

Multiply and divide the bracket by 99:

Sn=79 (9+99+999+⋯(n terms)).S_n=\dfrac79\,(9+99+999+\cdots\text{(}n\text{ terms)}).

Each term inside is one less than a power of 1010: 9=10−1, 99=102−1, 999=103−19=10-1,\ 99=10^2-1,\ 999=10^3-1, etc. So

Sn=79[(10+102+103+⋯+10n)−n].S_n=\dfrac79\Big[(10+10^2+10^3+\cdots+10^n)-n\Big].

The bracketed sum is a G.P. with first term 1010, ratio 1010, and nn terms:

10+102+⋯+10n=10(10n−1)10−1=10(10n−1)9=10n+1−109.10+10^2+\cdots+10^n=\dfrac{10(10^n-1)}{10-1}=\dfrac{10(10^n-1)}{9}=\dfrac{10^{n+1}-10}{9}.

So: …

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