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Question Number 125266 by ZiYangLee last updated on 09/Dec/20

$$\mathrm{Let}{S}_n=1+11+111+1111+11111+\dots \mathrm{to}n\mathrm{terms}$$$$\mathrm{Show}\mathrm{that}{S}_n=\frac{{10}^{n+1}-10-9n}{81}$$

Answered by Dwaipayan Shikari last updated on 09/Dec/20

$$S_n=\frac{1}{9}(9+99+999+9999+...)$$$$=\frac{1}{9}(10-1+100-1+1000-1+..+{10}^n-1)$$$$=\frac{1}{9}(10+{10}^2+{10}^3+..+{10}^n)-\frac{n}{9}$$$$=\frac{1}{9}(10.\frac{{10}^n-1}{9})-\frac{n}{9}=\frac{{10}^{n+1}-10-9n}{81}$$

Commented by Dwaipayan Shikari last updated on 09/Dec/20

$$Thereare\frac{1}{9}(10-1+100-1+1000-1+...+{10}^n-1)$$$$Means\frac{1}{9}(10+{10}^2+{10}^3+..+{10}^n)-\frac{1}{9}\left(n\right)$$

Commented by ZiYangLee last updated on 09/Dec/20

$$\mathrm{oh}\mathrm{I}\mathrm{see}!\mathrm{Thanks}\mathrm{very}\mathrm{much}!$$

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