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23-2323-232323-n-terms-

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Question Number 175457 by Rasheed.Sindhi last updated on 30/Aug/22
23+2323+232323+....(n terms)=?
$$ \\ $$$$\mathrm{23}+\mathrm{2323}+\mathrm{232323}+….\left({n}\:{terms}\right)=? \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 01/Sep/22
AnOther way... 23_(+2323_(+232323_(+23232323_(+2323232323) ) ) ) ................. 23+2300+230000+...n terms 23+2300+230000+...n-1 terms 23+2300+230000+...n-2 terms ................................................ ................................................ 23+2300.......................2 terms 23.................................1 term ((23(100^n −1))/(100−1))+((23(100^(n−1) −1))/(100−1))+((23(100^(n−2) −1))/(100−1))+...+((23(100^2 −1))/(100−1))+((23(100^1 −1))/(100−1)) (1/(99)){(23∙100^n +23∙100^(n−1) +23∙100^(n−2) +...n terms)−23n} (1/(99)){(((23∙100^n (((1/(100)))^n −1))/((1/(100))−1)))−23n} (1/(99)){(((23−23∙100^n )/((−99)/(100))))−23n} (1/(99)){((23−23∙100^n )/(−99))∙100}−((23n)/(99)) (1/(99)){((23∙100^(n+1) −23∙100)/(99))}−((23n)/(99)) ((2300)/(99)){((100^n −1)/(99))}−((23n)/(99)) ((2300(100^n −1)−2277n)/(9801)) ((23(100^(n+1) −99n−100))/(9801))
$$\boldsymbol{\mathrm{AnOther}}\:\boldsymbol{\mathrm{way}}… \\ $$$$\underset{\underset{\underset{\underset{+\mathrm{2323232323}} {+\mathrm{23232323}}} {+\mathrm{232323}}} {+\mathrm{2323}}} {\:\:\:\:\mathrm{23}} \\ $$$$…………….. \\ $$$$\mathrm{23}+\mathrm{2300}+\mathrm{230000}+…{n}\:{terms} \\ $$$$\mathrm{23}+\mathrm{2300}+\mathrm{230000}+…{n}-\mathrm{1}\:{terms} \\ $$$$\mathrm{23}+\mathrm{2300}+\mathrm{230000}+…{n}-\mathrm{2}\:{terms} \\ $$$$………………………………………… \\ $$$$………………………………………… \\ $$$$\mathrm{23}+\mathrm{2300}…………………..\mathrm{2}\:{terms} \\ $$$$\mathrm{23}……………………………\mathrm{1}\:{term} \\ $$$$\frac{\mathrm{23}\left(\mathrm{100}^{{n}} −\mathrm{1}\right)}{\mathrm{100}−\mathrm{1}}+\frac{\mathrm{23}\left(\mathrm{100}^{{n}−\mathrm{1}} −\mathrm{1}\right)}{\mathrm{100}−\mathrm{1}}+\frac{\mathrm{23}\left(\mathrm{100}^{{n}−\mathrm{2}} −\mathrm{1}\right)}{\mathrm{100}−\mathrm{1}}+…+\frac{\mathrm{23}\left(\mathrm{100}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{100}−\mathrm{1}}+\frac{\mathrm{23}\left(\mathrm{100}^{\mathrm{1}} −\mathrm{1}\right)}{\mathrm{100}−\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{99}}\left\{\left(\mathrm{23}\centerdot\mathrm{100}^{{n}} +\mathrm{23}\centerdot\mathrm{100}^{{n}−\mathrm{1}} +\mathrm{23}\centerdot\mathrm{100}^{{n}−\mathrm{2}} +…{n}\:{terms}\right)−\mathrm{23}{n}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{99}}\left\{\left(\frac{\mathrm{23}\centerdot\mathrm{100}^{{n}} \left(\left(\frac{\mathrm{1}}{\mathrm{100}}\right)^{{n}} −\mathrm{1}\right)}{\frac{\mathrm{1}}{\mathrm{100}}−\mathrm{1}}\right)−\mathrm{23}{n}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{99}}\left\{\left(\frac{\mathrm{23}−\mathrm{23}\centerdot\mathrm{100}^{{n}} }{\frac{−\mathrm{99}}{\mathrm{100}}}\right)−\mathrm{23}{n}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{99}}\left\{\frac{\mathrm{23}−\mathrm{23}\centerdot\mathrm{100}^{{n}} }{−\mathrm{99}}\centerdot\mathrm{100}\right\}−\frac{\mathrm{23}{n}}{\mathrm{99}} \\ $$$$\frac{\mathrm{1}}{\mathrm{99}}\left\{\frac{\mathrm{23}\centerdot\mathrm{100}^{{n}+\mathrm{1}} −\mathrm{23}\centerdot\mathrm{100}}{\mathrm{99}}\right\}−\frac{\mathrm{23}{n}}{\mathrm{99}} \\ $$$$\frac{\mathrm{2300}}{\mathrm{99}}\left\{\frac{\mathrm{100}^{{n}} −\mathrm{1}}{\mathrm{99}}\right\}−\frac{\mathrm{23}{n}}{\mathrm{99}} \\ $$$$\frac{\mathrm{2300}\left(\mathrm{100}^{{n}} −\mathrm{1}\right)−\mathrm{2277}{n}}{\mathrm{9801}} \\ $$$$\frac{\mathrm{23}\left(\mathrm{100}^{{n}+\mathrm{1}} −\mathrm{99}{n}−\mathrm{100}\right)}{\mathrm{9801}} \\ $$
Commented by infinityaction last updated on 01/Sep/22
amazing solution sir
$${amazing}\:{solution}\:{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Sep/22
Thanks sir!
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{sir}}! \\ $$
Commented by Tawa11 last updated on 15/Sep/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Sep/22
T^(H^A N) X_(MISS!)
$$\underset{\mathcal{MISS}!} {\mathbb{T}^{\mathbb{H}^{\mathbb{A}} \mathbb{N}} \mathbb{X}} \\ $$
Answered by Ar Brandon last updated on 30/Aug/22
Σu_n =23+2323+232323+23232323+... Δu_n =2300+230000+23000000+..., r=100 ⇒u_n =a×100^(n−1) +a_0 u_1 =23=a+a_0 u_2 =2323=100a+a_0 ⇒ { ((a+a_0 =23)),((100a+a_0 =2323)) :} ⇒a=((2300)/(99)), a_0 =23−((2300)/(99))=−((23)/(99)) ⇒u_n =((2300)/(99))×100^(n−1) −((23)/(99))=((23)/(99))(100^n −1) ⇒Σu_n =((2300)/(99))(((100^n −1)/(99)))−((23n)/(99)) =((23.100^(n+1) −2277n−2300)/(9801))
$$\Sigma{u}_{{n}} =\mathrm{23}+\mathrm{2323}+\mathrm{232323}+\mathrm{23232323}+… \\ $$$$\Delta{u}_{{n}} =\mathrm{2300}+\mathrm{230000}+\mathrm{23000000}+…,\:{r}=\mathrm{100} \\ $$$$\Rightarrow{u}_{{n}} ={a}×\mathrm{100}^{{n}−\mathrm{1}} +{a}_{\mathrm{0}} \\ $$$${u}_{\mathrm{1}} =\mathrm{23}={a}+{a}_{\mathrm{0}} \\ $$$${u}_{\mathrm{2}} =\mathrm{2323}=\mathrm{100}{a}+{a}_{\mathrm{0}} \:\Rightarrow\begin{cases}{{a}+{a}_{\mathrm{0}} =\mathrm{23}}\\{\mathrm{100}{a}+{a}_{\mathrm{0}} =\mathrm{2323}}\end{cases} \\ $$$$\Rightarrow{a}=\frac{\mathrm{2300}}{\mathrm{99}},\:{a}_{\mathrm{0}} =\mathrm{23}−\frac{\mathrm{2300}}{\mathrm{99}}=−\frac{\mathrm{23}}{\mathrm{99}} \\ $$$$\Rightarrow{u}_{{n}} =\frac{\mathrm{2300}}{\mathrm{99}}×\mathrm{100}^{{n}−\mathrm{1}} −\frac{\mathrm{23}}{\mathrm{99}}=\frac{\mathrm{23}}{\mathrm{99}}\left(\mathrm{100}^{{n}} −\mathrm{1}\right) \\ $$$$\Rightarrow\Sigma{u}_{{n}} =\frac{\mathrm{2300}}{\mathrm{99}}\left(\frac{\mathrm{100}^{{n}} −\mathrm{1}}{\mathrm{99}}\right)−\frac{\mathrm{23}{n}}{\mathrm{99}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{23}.\mathrm{100}^{{n}+\mathrm{1}} −\mathrm{2277}{n}−\mathrm{2300}}{\mathrm{9801}} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Sep/22
Thanks sir!
$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{sir}! \\ $$
Answered by Rasheed.Sindhi last updated on 31/Aug/22
S_n =23+2323+232323+....(n terms) (S_n /(23))=1+101+10101+1010101+...(n terms) ((99S_n )/(23))=99+9999+999999+99999999+...(n terms) =(100−1)+(10000−1)+(1000000−1)+...(n terms) =(10^2 +10^4 +10^6 +...(n terms))−n =((10^2 (10^(2n) −1))/(10^2 −1))−n =((10^(2n+2) −99n−10^2 )/(99)) S_n =((23(10^(2n+2) −99n−100))/(99×99)) =((23(10^(2n+2) −99n−100))/(9801))
$${S}_{{n}} =\mathrm{23}+\mathrm{2323}+\mathrm{232323}+….\left({n}\:{terms}\right) \\ $$$$\frac{{S}_{{n}} }{\mathrm{23}}=\mathrm{1}+\mathrm{101}+\mathrm{10101}+\mathrm{1010101}+…\left({n}\:{terms}\right) \\ $$$$\frac{\mathrm{99}{S}_{{n}} }{\mathrm{23}}=\mathrm{99}+\mathrm{9999}+\mathrm{999999}+\mathrm{99999999}+…\left({n}\:{terms}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{100}−\mathrm{1}\right)+\left(\mathrm{10000}−\mathrm{1}\right)+\left(\mathrm{1000000}−\mathrm{1}\right)+…\left({n}\:{terms}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{4}} +\mathrm{10}^{\mathrm{6}} +…\left({n}\:{terms}\right)\right)−{n} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{10}^{\mathrm{2}} \left(\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}\right)}{\mathrm{10}^{\mathrm{2}} −\mathrm{1}}−{n} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{10}^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{99}{n}−\mathrm{10}^{\mathrm{2}} }{\mathrm{99}} \\ $$$${S}_{{n}} =\frac{\mathrm{23}\left(\mathrm{10}^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{99}{n}−\mathrm{100}\right)}{\mathrm{99}×\mathrm{99}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{23}\left(\mathrm{10}^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{99}{n}−\mathrm{100}\right)}{\mathrm{9801}} \\ $$
Commented by peter frank last updated on 31/Aug/22
thanks
$$\mathrm{thanks} \\ $$

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