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Worked Examples · Example 22

Q.Find the sum of the sequence 4,44,444,…4, 44, 444, \ldots to nn terms.

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Factor out 4, rewrite each term as 11…1⏟k=10k−19\underbrace{11\ldots1}_{k}=\dfrac{10^k-1}{9}, then sum the resulting geometric series.

[!FORMULA] 11…1⏟k ones=10k−19\underbrace{11\ldots1}_{k\text{ ones}}=\dfrac{10^k-1}{9}, and ∑k=1n10k=10(10n−1)10−1=10n+1−109\displaystyle\sum_{k=1}^n 10^k=\dfrac{10(10^n-1)}{10-1}=\dfrac{10^{n+1}-10}{9}.

  1. Write 4+44+444+⋯  (n terms)=4(1+11+111+⋯+11…1⏟n)4+44+444+\cdots\;(n\text{ terms}) = 4\left(1+11+111+\cdots+\underbrace{11\ldots1}_{n}\right).
  2. Multiply and divide the bracket by 9: =49(9+99+999+⋯+99…9⏟n)=\dfrac{4}{9}\left(9+99+999+\cdots+\underbrace{99\ldots9}_{n}\right).
  3. Each bracket term is 10k−110^k-1, so =49[(10+102+⋯+10n)−n]=\dfrac{4}{9}\left[(10+10^2+\cdots+10^n)-n\right].
  4. Sum the G.P. 10+102+⋯+10n10+10^2+\cdots+10^n: =10(10n−1)10−1=10n+1−109=\dfrac{10(10^n-1)}{10-1}=\dfrac{10^{n+1}-10}{9}. …

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