7+67+667+6667+.....(n terms)=?


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

$7 + 67 + 667 + 6667 + . . . . . (n t e r m s) = ?$
Commented by infinityaction last updated on 30/Aug/22

$(6 + 1) + (66 + 1) + (666 + 1) + . . . (n t e r m s)$ $6 + 66 + . . . + n t e r m s + 1 + 1 + 1 . . . n t e r m s$ $6 [1 + 11 + 111 + . . + 1 (n t e r m s)] + n (1)$ $S_{n} = 1 + 11 + 111 + . . . + 1111 . . . 11$ $9 S_{n} = 9 + 99 + 999 + . . . + 999 . . . 99$ $= (10 - 1) + (100 - 1) + (1000 - 1) + . . . n t e r m s$ $= (10 + 100 + 1000 + . . . n t e r m s) - n$ $= \frac{10 (10^{n} - 1)}{10 - 1} - n$ $= \frac{10 (10^{n} - 1) - 9 n}{9}$ $S_{n} = \frac{10^{n + 1} - 9 n - 10}{81}$ $t h e n b y {e q}^{n} (1)$ $6 [\frac{10^{n + 1} - 9 n - 10}{81}] + n$ $\frac{2}{27} [10^{n + 1} - 9 n - 10] + n$ $\frac{2 \times 10^{n + 1} - 18 n + 27 n - 20}{27}$ $\frac{2 \times 10^{n + 1} + 9 n - 20}{27}$
Commented by Rasheed.Sindhi last updated on 30/Aug/22

$T han k s to all sirs for their good &$ $unique approaches!$
Commented by peter frank last updated on 31/Aug/22

$thanks$
	Answered by Ar Brandon last updated on 29/Aug/22
	$u_n =7+67+667+6667+... Δu_n =60+600+6000, r=10 ⇒u_n =a×10^(n−1) +a_0 u_1 =7=a+a_0 u_2 =67=10a+a_0 ⇒ { ((a+a_0 =7)),((10a+a_0 =67)) :} ⇒a=((60)/9) , a_0 =(1/3) ⇒u_n =((20)/3)×10^(n−1) +(1/3) Σu_n =((20)/3)(((10^n −1)/9))+(n/3)$
	$u_{n} = 7 + 67 + 667 + 6667 + . . .$ $Δ u_{n} = 60 + 600 + 6000, r = 10$ $\Rightarrow u_{n} = a \times 10^{n - 1} + a_{0}$ $u_{1} = 7 = a + a_{0}$ $u_{2} = 67 = 10 a + a_{0} \Rightarrow {\begin{cases} a + a_{0} = 7 \\ 10 a + a_{0} = 67 \end{cases}$ $\Rightarrow a = \frac{60}{9}, a_{0} = \frac{1}{3}$ $\Rightarrow u_{n} = \frac{20}{3} \times 10^{n - 1} + \frac{1}{3}$ $Σ u_{n} = \frac{20}{3} (\frac{10^{n} - 1}{9}) + \frac{n}{3}$
	Answered by Rasheed.Sindhi last updated on 30/Aug/22
	$S=7+67+667+6667+.....(n terms) S=(10−3)+(100−33)+(1000−333)+...(n terms) ={10+100+100+...(n terms)}_(S1) −{3+33+333+...(n terms)} =S_1 −(3/9)(9+99+999+...(n terms) =S_1 −(1/3)(10−1+100−1+1000−1+...(n terms) =S_1 −(1/3){10+100+1000+...(n terms)−n} =S_1 −(1/3){S_1 −n} =(2/3)S_1 +(1/3)n ∵ S_1 =((10(10^n −1))/(10−1))=((10^(n+1) −10)/9) ∴S_n =(2/3)(((10^(n+1) −10)/9))+(1/3)n =((2∙10^(n+1) −20)/(27))+(n/3) =((2∙10^(n+1) +9n−20)/(27))$
	$S = 7 + 67 + 667 + 6667 + . . . . . (n t e r m s)$ $S = (10 - 3) + (100 - 33) + (1000 - 333) + . . . (n t e r m s)$ $= \underset{S 1}{\underset{⏟}{{10 + 100 + 100 + . . . (n t e r m s)}}}$ $- {3 + 33 + 333 + . . . (n t e r m s)}$ $= S_{1} - \frac{3}{9} (9 + 99 + 999 + . . . (n t e r m s)$ $= S_{1} - \frac{1}{3} (10 - 1 + 100 - 1 + 1000 - 1 + . . . (n t e r m s)$ $= S_{1} - \frac{1}{3} {10 + 100 + 1000 + . . . (n t e r m s) - n}$ $= S_{1} - \frac{1}{3} {S_{1} - n}$ $= \frac{2}{3} S_{1} + \frac{1}{3} n$ $∵ S_{1} = \frac{10 (10^{n} - 1)}{10 - 1} = \frac{10^{n + 1} - 10}{9}$ $∴ S_{n} = \frac{2}{3} (\frac{10^{n + 1} - 10}{9}) + \frac{1}{3} n$ $= \frac{2 \cdot 10^{n + 1} - 20}{27} + \frac{n}{3}$ $= \frac{2 \cdot 10^{n + 1} + 9 n - 20}{27}$
	Commented by Tawa11 last updated on 30/Aug/22

	$Great sirs$
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