23+2323+232323+....(n terms)=?


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Question Number 175457 by Rasheed.Sindhi last updated on 30/Aug/22

$23 + 2323 + 232323 + . . . . (n t e r m s) = ?$
	Answered by Rasheed.Sindhi last updated on 01/Sep/22
	$AnOther way... 23_(+2323_(+232323_(+23232323_(+2323232323) ) ) ) ................. 23+2300+230000+...n terms 23+2300+230000+...n-1 terms 23+2300+230000+...n-2 terms ................................................ ................................................ 23+2300.......................2 terms 23.................................1 term ((23(100^n −1))/(100−1))+((23(100^(n−1) −1))/(100−1))+((23(100^(n−2) −1))/(100−1))+...+((23(100^2 −1))/(100−1))+((23(100^1 −1))/(100−1)) (1/(99)){(23∙100^n +23∙100^(n−1) +23∙100^(n−2) +...n terms)−23n} (1/(99)){(((23∙100^n (((1/(100)))^n −1))/((1/(100))−1)))−23n} (1/(99)){(((23−23∙100^n )/((−99)/(100))))−23n} (1/(99)){((23−23∙100^n )/(−99))∙100}−((23n)/(99)) (1/(99)){((23∙100^(n+1) −23∙100)/(99))}−((23n)/(99)) ((2300)/(99)){((100^n −1)/(99))}−((23n)/(99)) ((2300(100^n −1)−2277n)/(9801)) ((23(100^(n+1) −99n−100))/(9801))$
	$AnOther way . . .$ $\underset{\underset{\underset{\underset{+ 2323232323}{+ 23232323}}{+ 232323}}{+ 2323}}{23}$ $. . . . . . . . . . . . . . . . .$ $23 + 2300 + 230000 + . . . n t e r m s$ $23 + 2300 + 230000 + . . . n - 1 t e r m s$ $23 + 2300 + 230000 + . . . n - 2 t e r m s$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $23 + 2300 . . . . . . . . . . . . . . . . . . . . . . . 2 t e r m s$ $23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 t e r m$ $\frac{23 (100^{n} - 1)}{100 - 1} + \frac{23 (100^{n - 1} - 1)}{100 - 1} + \frac{23 (100^{n - 2} - 1)}{100 - 1} + . . . + \frac{23 (100^{2} - 1)}{100 - 1} + \frac{23 (100^{1} - 1)}{100 - 1}$ $\frac{1}{99} {(23 \cdot 100^{n} + 23 \cdot 100^{n - 1} + 23 \cdot 100^{n - 2} + . . . n t e r m s) - 23 n}$ $\frac{1}{99} {(\frac{23 \cdot 100^{n} ({(\frac{1}{100})}^{n} - 1)}{\frac{1}{100} - 1}) - 23 n}$ $\frac{1}{99} {(\frac{23 - 23 \cdot 100^{n}}{\frac{- 99}{100}}) - 23 n}$ $\frac{1}{99} {\frac{23 - 23 \cdot 100^{n}}{- 99} \cdot 100} - \frac{23 n}{99}$ $\frac{1}{99} {\frac{23 \cdot 100^{n + 1} - 23 \cdot 100}{99}} - \frac{23 n}{99}$ $\frac{2300}{99} {\frac{100^{n} - 1}{99}} - \frac{23 n}{99}$ $\frac{2300 (100^{n} - 1) - 2277 n}{9801}$ $\frac{23 (100^{n + 1} - 99 n - 100)}{9801}$
	Commented by infinityaction last updated on 01/Sep/22

	$a m a z i n g s o l u t i o n s i r$
	Commented by Rasheed.Sindhi last updated on 01/Sep/22

	$T han k s sir!$
	Commented by Tawa11 last updated on 15/Sep/22

	$Great sir$
	Commented by Rasheed.Sindhi last updated on 15/Sep/22

	$\underset{MISS!}{T^{H^{A} N} X}$
	Answered by Ar Brandon last updated on 30/Aug/22
	$Σu_n =23+2323+232323+23232323+... Δu_n =2300+230000+23000000+..., r=100 ⇒u_n =a×100^(n−1) +a_0 u_1 =23=a+a_0 u_2 =2323=100a+a_0 ⇒ { ((a+a_0 =23)),((100a+a_0 =2323)) :} ⇒a=((2300)/(99)), a_0 =23−((2300)/(99))=−((23)/(99)) ⇒u_n =((2300)/(99))×100^(n−1) −((23)/(99))=((23)/(99))(100^n −1) ⇒Σu_n =((2300)/(99))(((100^n −1)/(99)))−((23n)/(99)) =((23.100^(n+1) −2277n−2300)/(9801))$
	$Σ u_{n} = 23 + 2323 + 232323 + 23232323 + . . .$ $Δ u_{n} = 2300 + 230000 + 23000000 + . . ., r = 100$ $\Rightarrow u_{n} = a \times 100^{n - 1} + a_{0}$ $u_{1} = 23 = a + a_{0}$ $u_{2} = 2323 = 100 a + a_{0} \Rightarrow {\begin{cases} a + a_{0} = 23 \\ 100 a + a_{0} = 2323 \end{cases}$ $\Rightarrow a = \frac{2300}{99}, a_{0} = 23 - \frac{2300}{99} = - \frac{23}{99}$ $\Rightarrow u_{n} = \frac{2300}{99} \times 100^{n - 1} - \frac{23}{99} = \frac{23}{99} (100^{n} - 1)$ $\Rightarrow Σ u_{n} = \frac{2300}{99} (\frac{100^{n} - 1}{99}) - \frac{23 n}{99}$ $= \frac{23 . 100^{n + 1} - 2277 n - 2300}{9801}$
	Commented by Rasheed.Sindhi last updated on 01/Sep/22

	$T han k s sir!$
	Answered by Rasheed.Sindhi last updated on 31/Aug/22

	$S_{n} = 23 + 2323 + 232323 + . . . . (n t e r m s)$ $\frac{S_{n}}{23} = 1 + 101 + 10101 + 1010101 + . . . (n t e r m s)$ $\frac{99 S_{n}}{23} = 99 + 9999 + 999999 + 99999999 + . . . (n t e r m s)$ $= (100 - 1) + (10000 - 1) + (1000000 - 1) + . . . (n t e r m s)$ $= (10^{2} + 10^{4} + 10^{6} + . . . (n t e r m s)) - n$ $= \frac{10^{2} (10^{2 n} - 1)}{10^{2} - 1} - n$ $= \frac{10^{2 n + 2} - 99 n - 10^{2}}{99}$ $S_{n} = \frac{23 (10^{2 n + 2} - 99 n - 100)}{99 \times 99}$ $= \frac{23 (10^{2 n + 2} - 99 n - 100)}{9801}$
	Commented by peter frank last updated on 31/Aug/22

	$thanks$
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