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Question Number 1839 by RasheedAhmad last updated on 11/Oct/15
Sum to n terms the series: 0∙3+0∙33+0∙333+...
$${Sum}\:{to}\:{n}\:{terms}\:{the}\:{series}: \\ $$$$\mathrm{0}\centerdot\mathrm{3}+\mathrm{0}\centerdot\mathrm{33}+\mathrm{0}\centerdot\mathrm{333}+… \\ $$$$ \\ $$
Answered by 112358 last updated on 12/Oct/15
We may begin by searching for the general term u_n of the series. u_1 =0.3=3×10^(−1) u_2 =0.33=0.3+0.03=u_1 +3×10^(−2) u_3 =0.333=0.33+0.003=u_2 +3×10^(−3) From the present pattern we propose then that u_(n+1) =u_n +3×10^(−n−1) (n≥1). We may rewrite this equation as u_(n+1) −u_n =3×10^(−n−1) =3(10^(−1) )^(n+1) . Writting the formula for n≥1: u_(n+1) −u_n =3(10^(−1) )^(n+1) u_n −u_(n−1) =3(10^(−1) )^n u_(n−1) −u_(n−2) =3(10^(−1) )^(n−1) u_(n−2) −u_(n−3) =3(10^(−1) )^(n−2) u_(n−3) −u_(n−4) =3(10^(−1) )^(n−3) ... u_4 −u_3 =3(10^(−1) )^4 u_3 −u_2 =3(10^(−1) )^3 u_2 −u_1 =3(10^(−1) )^2 . Adding these equations together leads to all terms for 2≤n<n+1 being cancelled so we get u_(n+1) −u_1 =3(10^(−1) )^(n+1) +3(10^(−1) )^n +...+3(10^(−1) )^3 +3(10^(−1) )^2 Thus, u_(n+1) =3(10)^(−1) +Σ_(r=2) ^n 3(10^(−1) )^r u_(n+1) =3Σ_(r=1) ^(n+1) (10^(−1) )^r =0.3Σ_(r=1) ^(n+1) (10^(−1) )^(r−1) The summation simplifies as follows. Σ_(r=1) ^(n+1) (10^(−1) )^(r−1) =1+10^(−1) +10^(−2) +...+10^(−n) =(((10^(−1) )^(n+1) −1)/(10^(−1) −1)) =((10^(n+1) −1)/(0.9×10^(n+1) )) ∴u_(n+1) =((0.3(10^(n+1) −1))/(0.9(10^(n+1) )))=((10^(n+1) −1)/(3(10^(n+1) ))) The general term is then of the form u_n =((10^n −1)/(3(10^n ))). Checking that this formular works try for n=1,2,3. n=1: u_1 =((10−1)/(3×10))=(9/(3×10))=(3/(10))=0.3 as given. n=2: u_2 =((100−1)/(3(100)))=((99)/(3(100)))=((33)/(100))=0.33 as given. n=3: u_3 =((1000−1)/(3×1000))=((999)/(3×1000))=((333)/(1000))=0.333 as given. We now go about finding the sum Σ_(r=1) ^n u_r . Define the required sum as S_n =Σ_(r=1) ^n u_r . Then S_n =Σ_(r=1) ^n ((10^r −1)/(3(10^r )))=(1/3)Σ_(r=1) ^n (1−(10^(−1) )^r ) S_n =(1/3)Σ_(r=1) ^n 1−((10^(−1) )/3)Σ_(r=1) ^n (10^(−1) )^(r−1) =(n/3)−(1/(30))(((10^(−n) −1)/(10^(−1) −1))) =(n/3)−(1/(30))×((−10)/9)((1/(10^n ))−1) =(n/3)+(1/(27))(((1−10^n )/(10^n ))) =((27n(10^n )+3−3(10^n ))/(81(10^n ))) S_n =(((9n−1)10^n +1)/(27(10^n ))). Let′s test the formular for S_n . n=1: S_1 =((8×10+1)/(27×10))=((81)/(27×10))=0.3 as given. n=2: S_2 =((17×100+1)/(27×100))=((1701)/(27×100))=((189)/(3×100))=((63)/(100))=0.63 as given. n=3: S_3 =((26×1000+1)/(27×1000))=((963)/(1000))=0.963 as given. One can go further on to proving the result by induction on n for P(k)⇒P(k+1). I ran out of space for the proof.
$${We}\:{may}\:{begin}\:{by}\:{searching}\:{for} \\ $$$${the}\:{general}\:{term}\:{u}_{{n}} \:{of}\:{the}\:{series}. \\ $$$${u}_{\mathrm{1}} =\mathrm{0}.\mathrm{3}=\mathrm{3}×\mathrm{10}^{−\mathrm{1}} \\ $$$${u}_{\mathrm{2}} =\mathrm{0}.\mathrm{33}=\mathrm{0}.\mathrm{3}+\mathrm{0}.\mathrm{03}={u}_{\mathrm{1}} +\mathrm{3}×\mathrm{10}^{−\mathrm{2}} \\ $$$${u}_{\mathrm{3}} =\mathrm{0}.\mathrm{333}=\mathrm{0}.\mathrm{33}+\mathrm{0}.\mathrm{003}={u}_{\mathrm{2}} +\mathrm{3}×\mathrm{10}^{−\mathrm{3}} \\ $$$${From}\:{the}\:{present}\:{pattern}\:{we} \\ $$$${propose}\:{then}\:{that}\: \\ $$$${u}_{{n}+\mathrm{1}} ={u}_{{n}} +\mathrm{3}×\mathrm{10}^{−{n}−\mathrm{1}} \:\left({n}\geqslant\mathrm{1}\right).\:{We} \\ $$$${may}\:{rewrite}\:{this}\:{equation}\:{as} \\ $$$${u}_{{n}+\mathrm{1}} −{u}_{{n}} =\mathrm{3}×\mathrm{10}^{−{n}−\mathrm{1}} =\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{n}+\mathrm{1}} . \\ $$$${Writting}\:{the}\:{formula}\:{for}\:{n}\geqslant\mathrm{1}: \\ $$$${u}_{{n}+\mathrm{1}} −{u}_{{n}} =\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{n}+\mathrm{1}} \\ $$$${u}_{{n}} −{u}_{{n}−\mathrm{1}} =\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{n}} \\ $$$${u}_{{n}−\mathrm{1}} −{u}_{{n}−\mathrm{2}} =\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{n}−\mathrm{1}} \\ $$$${u}_{{n}−\mathrm{2}} −{u}_{{n}−\mathrm{3}} =\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{n}−\mathrm{2}} \\ $$$${u}_{{n}−\mathrm{3}} −{u}_{{n}−\mathrm{4}} =\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{n}−\mathrm{3}} \\ $$$$… \\ $$$${u}_{\mathrm{4}} −{u}_{\mathrm{3}} =\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{\mathrm{4}} \\ $$$${u}_{\mathrm{3}} −{u}_{\mathrm{2}} =\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{\mathrm{3}} \\ $$$${u}_{\mathrm{2}} −{u}_{\mathrm{1}} =\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{\mathrm{2}} . \\ $$$${Adding}\:{these}\:{equations}\:{together} \\ $$$${leads}\:{to}\:{all}\:{terms}\:{for}\:\mathrm{2}\leqslant{n}<{n}+\mathrm{1} \\ $$$${being}\:{cancelled}\:{so}\:{we}\:{get} \\ $$$${u}_{{n}+\mathrm{1}} −{u}_{\mathrm{1}} =\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{n}+\mathrm{1}} +\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{n}} +…+\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{\mathrm{2}} \\ $$$${Thus}, \\ $$$${u}_{{n}+\mathrm{1}} =\mathrm{3}\left(\mathrm{10}\right)^{−\mathrm{1}} +\underset{{r}=\mathrm{2}} {\overset{{n}} {\sum}}\mathrm{3}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{r}} \\ $$$${u}_{{n}+\mathrm{1}} =\mathrm{3}\underset{{r}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{r}} =\mathrm{0}.\mathrm{3}\underset{{r}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{r}−\mathrm{1}} \\ $$$${The}\:{summation}\:{simplifies}\:{as} \\ $$$${follows}. \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{r}−\mathrm{1}} =\mathrm{1}+\mathrm{10}^{−\mathrm{1}} +\mathrm{10}^{−\mathrm{2}} +…+\mathrm{10}^{−{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{10}^{−\mathrm{1}} \right)^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{10}^{−\mathrm{1}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{0}.\mathrm{9}×\mathrm{10}^{{n}+\mathrm{1}} } \\ $$$$\therefore{u}_{{n}+\mathrm{1}} =\frac{\mathrm{0}.\mathrm{3}\left(\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{1}\right)}{\mathrm{0}.\mathrm{9}\left(\mathrm{10}^{{n}+\mathrm{1}} \right)}=\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{1}}{\mathrm{3}\left(\mathrm{10}^{{n}+\mathrm{1}} \right)} \\ $$$${The}\:{general}\:{term}\:{is}\:{then}\:{of}\:{the} \\ $$$${form}\:{u}_{{n}} =\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{3}\left(\mathrm{10}^{{n}} \right)}.\:{Checking}\:{that} \\ $$$${this}\:{formular}\:{works}\:{try}\:{for}\:{n}=\mathrm{1},\mathrm{2},\mathrm{3}. \\ $$$${n}=\mathrm{1}:\:{u}_{\mathrm{1}} =\frac{\mathrm{10}−\mathrm{1}}{\mathrm{3}×\mathrm{10}}=\frac{\mathrm{9}}{\mathrm{3}×\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{10}}=\mathrm{0}.\mathrm{3}\:\:{as}\:{given}. \\ $$$${n}=\mathrm{2}:\:{u}_{\mathrm{2}} =\frac{\mathrm{100}−\mathrm{1}}{\mathrm{3}\left(\mathrm{100}\right)}=\frac{\mathrm{99}}{\mathrm{3}\left(\mathrm{100}\right)}=\frac{\mathrm{33}}{\mathrm{100}}=\mathrm{0}.\mathrm{33}\:\:{as}\:{given}. \\ $$$${n}=\mathrm{3}:\:{u}_{\mathrm{3}} =\frac{\mathrm{1000}−\mathrm{1}}{\mathrm{3}×\mathrm{1000}}=\frac{\mathrm{999}}{\mathrm{3}×\mathrm{1000}}=\frac{\mathrm{333}}{\mathrm{1000}}=\mathrm{0}.\mathrm{333}\:\:{as}\:{given}. \\ $$$${We}\:{now}\:{go}\:{about}\:{finding}\:{the}\:{sum} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{u}_{{r}} .\:{Define}\:{the}\:{required}\:{sum}\:{as} \\ $$$${S}_{{n}} =\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{u}_{{r}} .\:{Then} \\ $$$${S}_{{n}} =\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{10}^{{r}} −\mathrm{1}}{\mathrm{3}\left(\mathrm{10}^{{r}} \right)}=\frac{\mathrm{1}}{\mathrm{3}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\left(\mathrm{10}^{−\mathrm{1}} \right)^{{r}} \right) \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1}−\frac{\mathrm{10}^{−\mathrm{1}} }{\mathrm{3}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{10}^{−\mathrm{1}} \right)^{{r}−\mathrm{1}} \\ $$$$\:\:\:\:\:=\frac{{n}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{30}}\left(\frac{\mathrm{10}^{−{n}} −\mathrm{1}}{\mathrm{10}^{−\mathrm{1}} −\mathrm{1}}\right) \\ $$$$\:\:\:\:=\frac{{n}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{30}}×\frac{−\mathrm{10}}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{10}^{{n}} }−\mathrm{1}\right) \\ $$$$\:\:\:=\frac{{n}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{27}}\left(\frac{\mathrm{1}−\mathrm{10}^{{n}} }{\mathrm{10}^{{n}} }\right) \\ $$$$\:\:\:=\frac{\mathrm{27}{n}\left(\mathrm{10}^{{n}} \right)+\mathrm{3}−\mathrm{3}\left(\mathrm{10}^{{n}} \right)}{\mathrm{81}\left(\mathrm{10}^{{n}} \right)} \\ $$$${S}_{{n}} =\frac{\left(\mathrm{9}{n}−\mathrm{1}\right)\mathrm{10}^{{n}} +\mathrm{1}}{\mathrm{27}\left(\mathrm{10}^{{n}} \right)}.\:{Let}'{s}\:{test}\:{the} \\ $$$${formular}\:{for}\:{S}_{{n}} . \\ $$$${n}=\mathrm{1}:\:{S}_{\mathrm{1}} =\frac{\mathrm{8}×\mathrm{10}+\mathrm{1}}{\mathrm{27}×\mathrm{10}}=\frac{\mathrm{81}}{\mathrm{27}×\mathrm{10}}=\mathrm{0}.\mathrm{3}\:\:{as}\:{given}. \\ $$$${n}=\mathrm{2}:\:{S}_{\mathrm{2}} =\frac{\mathrm{17}×\mathrm{100}+\mathrm{1}}{\mathrm{27}×\mathrm{100}}=\frac{\mathrm{1701}}{\mathrm{27}×\mathrm{100}}=\frac{\mathrm{189}}{\mathrm{3}×\mathrm{100}}=\frac{\mathrm{63}}{\mathrm{100}}=\mathrm{0}.\mathrm{63}\:\:{as}\:{given}. \\ $$$${n}=\mathrm{3}:\:{S}_{\mathrm{3}} =\frac{\mathrm{26}×\mathrm{1000}+\mathrm{1}}{\mathrm{27}×\mathrm{1000}}=\frac{\mathrm{963}}{\mathrm{1000}}=\mathrm{0}.\mathrm{963}\:\:{as}\:{given}. \\ $$$${One}\:{can}\:{go}\:{further}\:{on}\:{to}\:{proving} \\ $$$${the}\:{result}\:{by}\:{induction}\:{on}\:{n}\:{for} \\ $$$${P}\left({k}\right)\Rightarrow{P}\left({k}+\mathrm{1}\right). \\ $$$${I}\:{ran}\:{out}\:{of}\:{space}\:{for}\:{the}\:{proof}. \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 12/Oct/15
An other method This is a general method for such series: 2+22+222+...,4+44+444+...etc. 0∙3+0∙33+0∙333+... 3(0∙1+0∙11+0∙111+...) Dividing and multiplying by 9 : (3/9)(0.9+0.99+0.999+...) (3/9){(1−0.1)+(1−0.01)+(1−0.001)+...} (3/9){(1+1+...n terms)−(0.1+0.01+...n terms)} Wrtten in red colour is geometric series of n terms having first term(a) 0.1 and common ratio(r) 0.1 S_n =((a(1−r^n ))/(1−r)) (1/3){n−((((0.1) [1−(0.1)^n ])/(1−0.1)))} (1/3){n−((0.1[1−(0.1)^n ])/(0.9))} (1/3){((9n−[1−((1/(10)))^n ])/9)} (1/(27)){9n−1+(1/(10^n ))}
$${An}\:{other}\:{method} \\ $$$${This}\:{is}\:{a}\:{general}\:{method}\:{for}\:{such}\:{series}: \\ $$$$\:\mathrm{2}+\mathrm{22}+\mathrm{222}+…,\mathrm{4}+\mathrm{44}+\mathrm{444}+…{etc}. \\ $$$$\mathrm{0}\centerdot\mathrm{3}+\mathrm{0}\centerdot\mathrm{33}+\mathrm{0}\centerdot\mathrm{333}+… \\ $$$$\mathrm{3}\left(\mathrm{0}\centerdot\mathrm{1}+\mathrm{0}\centerdot\mathrm{11}+\mathrm{0}\centerdot\mathrm{111}+…\right) \\ $$$${Dividing}\:{and}\:{multiplying}\:{by}\:\mathrm{9}\:: \\ $$$$\frac{\mathrm{3}}{\mathrm{9}}\left(\mathrm{0}.\mathrm{9}+\mathrm{0}.\mathrm{99}+\mathrm{0}.\mathrm{999}+…\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{9}}\left\{\left(\mathrm{1}−\mathrm{0}.\mathrm{1}\right)+\left(\mathrm{1}−\mathrm{0}.\mathrm{01}\right)+\left(\mathrm{1}−\mathrm{0}.\mathrm{001}\right)+…\right\} \\ $$$$\frac{\mathrm{3}}{\mathrm{9}}\left\{\left(\mathrm{1}+\mathrm{1}+…{n}\:{terms}\right)−\left(\mathrm{0}.\mathrm{1}+\mathrm{0}.\mathrm{01}+…\mathrm{n}\:\mathrm{terms}\right)\right\} \\ $$$$\mathrm{Wrtten}\:\mathrm{in}\:\mathrm{red}\:\mathrm{colour}\:\mathrm{is}\:\mathrm{geometric}\:\mathrm{series} \\ $$$$\mathrm{of}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{having}\:\mathrm{first}\:\mathrm{term}\left({a}\right)\:\mathrm{0}.\mathrm{1}\:\mathrm{and}\:\mathrm{common} \\ $$$$\mathrm{ratio}\left({r}\right)\:\mathrm{0}.\mathrm{1} \\ $$$$\mathrm{S}_{\mathrm{n}} =\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left\{{n}−\left(\frac{\left(\mathrm{0}.\mathrm{1}\right)\:\left[\mathrm{1}−\left(\mathrm{0}.\mathrm{1}\right)^{\mathrm{n}} \right]}{\mathrm{1}−\mathrm{0}.\mathrm{1}}\right)\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left\{{n}−\frac{\mathrm{0}.\mathrm{1}\left[\mathrm{1}−\left(\mathrm{0}.\mathrm{1}\right)^{{n}} \right]}{\mathrm{0}.\mathrm{9}}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left\{\frac{\mathrm{9}{n}−\left[\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{10}}\right)^{{n}} \right]}{\mathrm{9}}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{27}}\left\{\mathrm{9}{n}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{10}^{{n}} }\right\} \\ $$$$ \\ $$

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