7-67-667-6667-n-terms-

Question Number 175411 by Rasheed.Sindhi last updated on 29/Aug/22

$$ \\ $$$$\mathrm{7}+\mathrm{67}+\mathrm{667}+\mathrm{6667}+…..\left({n}\:{terms}\right)=? \\ $$$$ \\ $$

Commented by infinityaction last updated on 30/Aug/22

(6+1)+(66+1)+(666+1)+...(n terms) 6+66+...+n terms + 1+1+1...n terms 6[1+11+111+..+1(n terms)]+n (1) S_n =1+11+111+...+1111...11 9S_n =9+99+999+...+999...99 =(10−1)+(100−1)+(1000−1)+... n terms =(10+100+1000+...n terms)−n =((10(10^n −1))/(10−1))−n =((10(10^n −1)−9n)/9) S_n =((10^(n+1) −9n−10)/(81)) then by eq^n (1) 6[((10^(n+1) −9n−10)/(81))]+n (2/(27))[10^(n+1) −9n−10]+n ((2×10^(n+1) −18n+27n−20)/(27)) ((2×10^(n+1) +9n−20)/(27))

$$\left(\mathrm{6}+\mathrm{1}\right)+\left(\mathrm{66}+\mathrm{1}\right)+\left(\mathrm{666}+\mathrm{1}\right)+…\left({n}\:{terms}\right) \\ $$$$\mathrm{6}+\mathrm{66}+…+{n}\:{terms}\:+\:\mathrm{1}+\mathrm{1}+\mathrm{1}…{n}\:{terms} \\ $$$$\mathrm{6}\left[\mathrm{1}+\mathrm{11}+\mathrm{111}+..+\mathrm{1}\left({n}\:{terms}\right)\right]+{n}\:\:\:\:\:\:\left(\mathrm{1}\right)\: \\ $$$$\:\:\:{S}_{{n}} =\mathrm{1}+\mathrm{11}+\mathrm{111}+…+\mathrm{1111}…\mathrm{11} \\ $$$$\mathrm{9}{S}_{{n}} =\mathrm{9}+\mathrm{99}+\mathrm{999}+…+\mathrm{999}…\mathrm{99} \\ $$$$\:\:=\left(\mathrm{10}−\mathrm{1}\right)+\left(\mathrm{100}−\mathrm{1}\right)+\left(\mathrm{1000}−\mathrm{1}\right)+…\:{n}\:{terms} \\ $$$$\:\:\:\:\:=\left(\mathrm{10}+\mathrm{100}+\mathrm{1000}+…{n}\:{terms}\right)−{n} \\ $$$$\:\:\:\:\:=\frac{\mathrm{10}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{10}−\mathrm{1}}−{n}\:\:\:\: \\ $$$$\:\:\:\:\:=\frac{\mathrm{10}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)−\mathrm{9}{n}}{\mathrm{9}} \\ $$$$\:\:\:\:{S}_{{n}} =\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{9}{n}−\mathrm{10}}{\mathrm{81}}\:\:\: \\ $$$$\:\:\:\:{then}\:\:\:\:\:\:{by}\:\:{eq}^{{n}} \left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\mathrm{6}\left[\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{9}{n}−\mathrm{10}}{\mathrm{81}}\right]+{n} \\ $$$$\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{27}}\left[\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{9}{n}−\mathrm{10}\right]+{n} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{2}×\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{18}{n}+\mathrm{27}{n}−\mathrm{20}}{\mathrm{27}} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{2}×\mathrm{10}^{{n}+\mathrm{1}} +\mathrm{9}{n}−\mathrm{20}}{\mathrm{27}} \\ $$

Commented by Rasheed.Sindhi last updated on 30/Aug/22

Thanks to all sirs for their good & unique approaches!

$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{sirs}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{their}}\:\boldsymbol{\mathrm{good}}\:\& \\ $$$$\boldsymbol{\mathrm{unique}}\:\boldsymbol{\mathrm{approaches}}! \\ $$

Commented by peter frank last updated on 31/Aug/22

$$\mathrm{thanks} \\ $$

Answered by Ar Brandon last updated on 29/Aug/22

u_n =7+67+667+6667+... Δu_n =60+600+6000, r=10 ⇒u_n =a×10^(n−1) +a_0 u_1 =7=a+a_0 u_2 =67=10a+a_0 ⇒ { ((a+a_0 =7)),((10a+a_0 =67)) :} ⇒a=((60)/9) , a_0 =(1/3) ⇒u_n =((20)/3)×10^(n−1) +(1/3) Σu_n =((20)/3)(((10^n −1)/9))+(n/3)

$${u}_{{n}} =\mathrm{7}+\mathrm{67}+\mathrm{667}+\mathrm{6667}+… \\ $$$$\Delta{u}_{{n}} =\mathrm{60}+\mathrm{600}+\mathrm{6000},\:{r}=\mathrm{10} \\ $$$$\Rightarrow{u}_{{n}} ={a}×\mathrm{10}^{{n}−\mathrm{1}} +{a}_{\mathrm{0}} \\ $$$${u}_{\mathrm{1}} =\mathrm{7}={a}+{a}_{\mathrm{0}} \\ $$$${u}_{\mathrm{2}} =\mathrm{67}=\mathrm{10}{a}+{a}_{\mathrm{0}} \:\Rightarrow\begin{cases}{{a}+{a}_{\mathrm{0}} =\mathrm{7}}\\{\mathrm{10}{a}+{a}_{\mathrm{0}} =\mathrm{67}}\end{cases} \\ $$$$\Rightarrow{a}=\frac{\mathrm{60}}{\mathrm{9}}\:,\:{a}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{u}_{{n}} =\frac{\mathrm{20}}{\mathrm{3}}×\mathrm{10}^{{n}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Sigma{u}_{{n}} =\frac{\mathrm{20}}{\mathrm{3}}\left(\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{9}}\right)+\frac{{n}}{\mathrm{3}} \\ $$

Answered by Rasheed.Sindhi last updated on 30/Aug/22

S=7+67+667+6667+.....(n terms) S=(10−3)+(100−33)+(1000−333)+...(n terms) ={10+100+100+...(n terms)}_(S1) −{3+33+333+...(n terms)} =S_1 −(3/9)(9+99+999+...(n terms) =S_1 −(1/3)(10−1+100−1+1000−1+...(n terms) =S_1 −(1/3){10+100+1000+...(n terms)−n} =S_1 −(1/3){S_1 −n} =(2/3)S_1 +(1/3)n ∵ S_1 =((10(10^n −1))/(10−1))=((10^(n+1) −10)/9) ∴S_n =(2/3)(((10^(n+1) −10)/9))+(1/3)n =((2∙10^(n+1) −20)/(27))+(n/3) =((2∙10^(n+1) +9n−20)/(27))

$$\mathrm{S}=\mathrm{7}+\mathrm{67}+\mathrm{667}+\mathrm{6667}+…..\left({n}\:{terms}\right) \\ $$$$\mathrm{S}=\left(\mathrm{10}−\mathrm{3}\right)+\left(\mathrm{100}−\mathrm{33}\right)+\left(\mathrm{1000}−\mathrm{333}\right)+…\left({n}\:{terms}\right) \\ $$$$=\underset{\mathrm{S1}} {\underbrace{\left\{\mathrm{10}+\mathrm{100}+\mathrm{100}+…\left({n}\:{terms}\right)\right\}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left\{\mathrm{3}+\mathrm{33}+\mathrm{333}+…\left({n}\:{terms}\right)\right\} \\ $$$$=\mathrm{S}_{\mathrm{1}} −\frac{\mathrm{3}}{\mathrm{9}}\left(\mathrm{9}+\mathrm{99}+\mathrm{999}+…\left({n}\:{terms}\right)\right. \\ $$$$=\mathrm{S}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{10}−\mathrm{1}+\mathrm{100}−\mathrm{1}+\mathrm{1000}−\mathrm{1}+…\left({n}\:{terms}\right)\right. \\ $$$$=\mathrm{S}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{3}}\left\{\mathrm{10}+\mathrm{100}+\mathrm{1000}+…\left({n}\:{terms}\right)−{n}\right\} \\ $$$$=\mathrm{S}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{3}}\left\{\mathrm{S}_{\mathrm{1}} −{n}\right\} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\mathrm{S}_{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{3}}{n} \\ $$$$\because\:\mathrm{S}_{\mathrm{1}} =\frac{\mathrm{10}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{10}−\mathrm{1}}=\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{10}}{\mathrm{9}} \\ $$$$\therefore\mathrm{S}_{{n}} =\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{10}}{\mathrm{9}}\right)+\frac{\mathrm{1}}{\mathrm{3}}{n} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}\centerdot\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{20}}{\mathrm{27}}+\frac{{n}}{\mathrm{3}} \\ $$$$\:\:\:\:=\frac{\mathrm{2}\centerdot\mathrm{10}^{{n}+\mathrm{1}} +\mathrm{9}{n}−\mathrm{20}}{\mathrm{27}} \\ $$

Commented by Tawa11 last updated on 30/Aug/22

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

Facebook Tweet Pin

7-67-667-6667-n-terms-

Leave a Reply Cancel reply