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Question Number 121235 by ZiYangLee last updated on 06/Nov/20

$$\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{of}n\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{series}$$$$S_n=1+22+333+4444+\dots \dots$$

Answered by Ar Brandon last updated on 06/Nov/20

$${\mathrm{S}}_{\mathrm{n}}=1+22+333+4444+\cdot \cdot \cdot$$$$=1\left({10}^0\right)+2({10}^1+{10}^0)+3({10}^2+{10}^1+{10}^0)$$$$+4({10}^3+{10}^2+{10}^1+{10}^0)$$$$={10}^0(1+2+3+\cdot \cdot \cdot +\mathrm{n})+{10}^1(2+3+4+\cdot \cdot \cdot +\mathrm{n})$$$$+{10}^2(3+4+5+\cdot \cdot \cdot +\mathrm{n})+\cdot \cdot \cdot +{10}^{\mathrm{n}-1}\left(\mathrm{n}\right)$$$$=\frac{\mathrm{n}(\mathrm{n}+1)}{2}+\frac{10(\mathrm{n}-1)(2+\mathrm{n})}{2}+\frac{{10}^2(\mathrm{n}-2)(3+\mathrm{n})}{2}+\cdot \cdot \cdot +\frac{{10}^{\mathrm{n}-1}\left(\mathrm{n}\right)}{1}$$$$$$$$Tobecontinued...$$

Answered by Dwaipayan Shikari last updated on 06/Nov/20

$$Firstterm=\frac{{10}^{n}n}{9}-\frac{n}{9}n=1$$$$secondterm=\frac{{10}^2.2}{9}-\frac{2}{9}$$$$thirdterm=\frac{{10}^3.3}{9}-\frac{3}{9}$$$$T_n=\frac{{10}^n.n}{9}-\frac{n}{9}$$$$\underset{n=1}{\sum ^n}{T}_n=\underset{n=1}{\sum ^n}\frac{{10}^n.n}{9}-\underset{n=1}{\sum ^n}\frac{n}{9}$$$$S=\sum ^{n}n.{10}^n$$$$S=10+2.{10}^2+...+n{10}^n$$$$-10S=-{10}^2-....-(n-1){10}^n-n{10}^{n+1}$$$$-9S=10+{10}^2+...+{10}^n-n{10}^{n+1}$$$$S=\frac{n}{9}{10}^{n+1}-\frac{1}{81}({10}^{n+1}-10)$$$$\sum ^n{T}_n=\frac{1}{9}(\frac{n}{9}{10}^{n+1}-\frac{1}{81}{10}^{n+1}+\frac{10}{81})-\sum ^n\frac{n}{9}$$$$=\frac{1}{9}(\frac{n}{9}{10}^{n+1}-\frac{1}{81}{10}^{n+1}+\frac{10}{81})-\frac{n(n+1)}{18}$$

Commented by Dwaipayan Shikari last updated on 06/Nov/20

$$Checking$$$$n=1$$$$\frac{1}{9}(\frac{100}{9}-\frac{100}{81}+\frac{10}{81})-\frac{2}{18}=\frac{10}{9}-\frac{1}{9}=1\left(true\right)$$$$$$

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