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

Let S_n =1+11+111+1111+11111+… to n terms  Show that S_n =((10^(n+1) −10−9n)/(81))

$$\mathrm{Let}\:{S}_{{n}} =\mathrm{1}+\mathrm{11}+\mathrm{111}+\mathrm{1111}+\mathrm{11111}+\ldots\:\mathrm{to}\:{n}\:\mathrm{terms} \\ $$$$\mathrm{Show}\:\mathrm{that}\:{S}_{{n}} =\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{10}−\mathrm{9}{n}}{\mathrm{81}} \\ $$

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

S_n =(1/9)(9+99+999+9999+...)  =(1/9)(10−1+100−1+1000−1+..+10^n −1)  =(1/9)(10+10^2 +10^3 +..+10^n )−(n/9)  =(1/9)(10.((10^n −1)/9))−(n/9)=((10^(n+1) −10−9n)/(81))

$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{9}+\mathrm{99}+\mathrm{999}+\mathrm{9999}+...\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{10}−\mathrm{1}+\mathrm{100}−\mathrm{1}+\mathrm{1000}−\mathrm{1}+..+\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{10}+\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{3}} +..+\mathrm{10}^{{n}} \right)−\frac{{n}}{\mathrm{9}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{10}.\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{9}}\right)−\frac{{n}}{\mathrm{9}}=\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{10}−\mathrm{9}{n}}{\mathrm{81}} \\ $$

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

There are (1/9)(10−1+100−1+1000−1+...+10^n −1)  Means (1/9)(10+10^2 +10^3 +..+10^n )−(1/9)(n)

$${There}\:{are}\:\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{10}−\mathrm{1}+\mathrm{100}−\mathrm{1}+\mathrm{1000}−\mathrm{1}+...+\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$$${Means}\:\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{10}+\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{3}} +..+\mathrm{10}^{{n}} \right)−\frac{\mathrm{1}}{\mathrm{9}}\left({n}\right) \\ $$

Commented by ZiYangLee last updated on 09/Dec/20

oh I see! Thanks very much!

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

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