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The sum is 1+11+111+1111+1111... find the sum of n

Guest Nov 24, 2015

Best Answer 

 #7
avatar+18712 
+11

The sum is 1+11+111+1111+1111... find the sum of n

 

\(\small{ \begin{array}{rcl} a_1 &=& 1 \\ a_2 &=& 11\\ a_3 &=& 111\\ a_4 &=& 1111\\ \dots \\ \end{array} \qquad \begin{array}{rcl} a_1 &=& 1 \\ a_2 &=& 1 + 10^1\\ a_3 &=& 1 + 10^1+ 10^2\\ a_4 &=& 1 + 10^1+ 10^2+10^3\\ \cdots \\ a_n &=& 1 + 10^1+ 10^2+10^3+10^4+\cdots + 10^{n-2}+ 10^{n-1}\\ \end{array}\\\\ \begin{array}{rcl} \\ S_n &=& ( n - 0 )\cdot 10^0 +( n - 1 )\cdot 10^1 +( n - 2 )\cdot 10^2 +( n - 3 )\cdot 10^3 +( n - 4 )\cdot 10^4\\ & + & \cdots +[ n - (n-2) ]\cdot 10^{n-2} +[ n - (n-1) ]\cdot 10^{n-1}\\ S_n &=& n\cdot 1 + n\cdot 10^1 + n \cdot 10^2 + n\cdot 10^4 + \cdots + n \cdot 10^{n-2} + n \cdot 10^{n-1} \\ & -& 0\cdot 10^0 - 1 \cdot 10^1 - 2\cdot 10^2 - 3\cdot 10^3 - 4\cdot 10^4 -\cdots -(n-2)\cdot 10^{n-2} -(n-1)\cdot 10^{n-1}\\ S_n &=& n\cdot \underbrace{( 1 + 10^1 + 10^2 + 10^4 + \cdots + 10^{n-2} + 10^{n-1} )}_{\text{geometric series}} - \underbrace{\sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} }}_{\text{'Arithmetic-geometric' series}}\\ S_n &=& n\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right) - \sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} }\\\\ S_n &=& n\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right) - s_n\\\\ \end{array}\\ \begin{array}{lclcl} \hline s_n = \sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} } &=& 0\cdot 10^0 + &1& \cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 +4\cdot 10^4 +\cdots + (n-1)\cdot 10^{n-1} \\ 10\cdot s_n &=& &0&\cdot 10^1+ 1\cdot 10^2 + 2\cdot 10^3 + 3\cdot 10^4 +\cdots + (n-2)\cdot 10^{n-1} + (n-1)\cdot 10^{n} \\ \hline \end{array}\\ \begin{array}{rcl} s_n- 10 s_n &=& \underbrace{1\cdot 10^1 + 1\cdot 10^2+1\cdot 10^3 +\cdots + 1\cdot 10^{n-2} + 1\cdot 10^{n-1}}_{\text{geometric series}} - (n-1)\cdot 10^n \\\\ -9 s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right)- (n-1)\cdot 10^n \\\\ -9 s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{9} \right)- (n-1)\cdot 10^n \\\\ - s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{9\cdot 9} \right)- \dfrac { (n-1)\cdot 10^n}{9} \\ \hline \end{array}\\ \begin{array}{rcl} S_n &=& n\cdot \left( \dfrac{10^{n-1}-1}{9} \right) - s_n\\\\ \end{array}\\\\ \begin{array}{rcl} \boxed{~ S_n = n\cdot \left( \dfrac{10^{n-1}-1}{9} \right) + 10\cdot \left( \dfrac{10^{n-1}-1}{9\cdot 9} \right)- \dfrac { (n-1)\cdot 10^n}{9} \\\\ ~} \end{array} }\)

 

laugh

heureka  Nov 24, 2015
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8+0 Answers

 #1
avatar+90988 
+5

The sum would be infinity     :D

Melody  Nov 24, 2015
 #2
avatar
+5

I meant find the sum of the equation, sorry

Guest Nov 24, 2015
 #3
avatar+90988 
0

There is no equation.

You have to have an equal sign for an equation.     frown

Melody  Nov 24, 2015
 #4
avatar+26322 
+10

The n'th term is given by \(x_n=10x_{n-1}+n\)

 

See below for some results:

sum of terms

Alan  Nov 24, 2015
 #5
avatar+90988 
0

Thanks Alan, I guess that is what was wanted......frown

Melody  Nov 24, 2015
 #6
avatar+26322 
+5

Possibly.  Certainly it was the only way I could make sense of the question!

Alan  Nov 24, 2015
 #7
avatar+18712 
+11
Best Answer

The sum is 1+11+111+1111+1111... find the sum of n

 

\(\small{ \begin{array}{rcl} a_1 &=& 1 \\ a_2 &=& 11\\ a_3 &=& 111\\ a_4 &=& 1111\\ \dots \\ \end{array} \qquad \begin{array}{rcl} a_1 &=& 1 \\ a_2 &=& 1 + 10^1\\ a_3 &=& 1 + 10^1+ 10^2\\ a_4 &=& 1 + 10^1+ 10^2+10^3\\ \cdots \\ a_n &=& 1 + 10^1+ 10^2+10^3+10^4+\cdots + 10^{n-2}+ 10^{n-1}\\ \end{array}\\\\ \begin{array}{rcl} \\ S_n &=& ( n - 0 )\cdot 10^0 +( n - 1 )\cdot 10^1 +( n - 2 )\cdot 10^2 +( n - 3 )\cdot 10^3 +( n - 4 )\cdot 10^4\\ & + & \cdots +[ n - (n-2) ]\cdot 10^{n-2} +[ n - (n-1) ]\cdot 10^{n-1}\\ S_n &=& n\cdot 1 + n\cdot 10^1 + n \cdot 10^2 + n\cdot 10^4 + \cdots + n \cdot 10^{n-2} + n \cdot 10^{n-1} \\ & -& 0\cdot 10^0 - 1 \cdot 10^1 - 2\cdot 10^2 - 3\cdot 10^3 - 4\cdot 10^4 -\cdots -(n-2)\cdot 10^{n-2} -(n-1)\cdot 10^{n-1}\\ S_n &=& n\cdot \underbrace{( 1 + 10^1 + 10^2 + 10^4 + \cdots + 10^{n-2} + 10^{n-1} )}_{\text{geometric series}} - \underbrace{\sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} }}_{\text{'Arithmetic-geometric' series}}\\ S_n &=& n\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right) - \sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} }\\\\ S_n &=& n\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right) - s_n\\\\ \end{array}\\ \begin{array}{lclcl} \hline s_n = \sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} } &=& 0\cdot 10^0 + &1& \cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 +4\cdot 10^4 +\cdots + (n-1)\cdot 10^{n-1} \\ 10\cdot s_n &=& &0&\cdot 10^1+ 1\cdot 10^2 + 2\cdot 10^3 + 3\cdot 10^4 +\cdots + (n-2)\cdot 10^{n-1} + (n-1)\cdot 10^{n} \\ \hline \end{array}\\ \begin{array}{rcl} s_n- 10 s_n &=& \underbrace{1\cdot 10^1 + 1\cdot 10^2+1\cdot 10^3 +\cdots + 1\cdot 10^{n-2} + 1\cdot 10^{n-1}}_{\text{geometric series}} - (n-1)\cdot 10^n \\\\ -9 s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right)- (n-1)\cdot 10^n \\\\ -9 s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{9} \right)- (n-1)\cdot 10^n \\\\ - s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{9\cdot 9} \right)- \dfrac { (n-1)\cdot 10^n}{9} \\ \hline \end{array}\\ \begin{array}{rcl} S_n &=& n\cdot \left( \dfrac{10^{n-1}-1}{9} \right) - s_n\\\\ \end{array}\\\\ \begin{array}{rcl} \boxed{~ S_n = n\cdot \left( \dfrac{10^{n-1}-1}{9} \right) + 10\cdot \left( \dfrac{10^{n-1}-1}{9\cdot 9} \right)- \dfrac { (n-1)\cdot 10^n}{9} \\\\ ~} \end{array} }\)

 

laugh

heureka  Nov 24, 2015
 #8
avatar+18712 
+10

The sum is 1+11+111+1111+1111... find the sum of n

 

Sorry blush

New edit, without mistakelaugh:

 

 

\(\small{ \begin{array}{rcl} a_1 &=& 1 \\ a_2 &=& 11\\ a_3 &=& 111\\ a_4 &=& 1111\\ \dots \\ \end{array} \qquad \begin{array}{rcl} a_1 &=& 1 \\ a_2 &=& 1 + 10^1\\ a_3 &=& 1 + 10^1+ 10^2\\ a_4 &=& 1 + 10^1+ 10^2+10^3\\ \cdots \\ a_n &=& 1 + 10^1+ 10^2+10^3+10^4+\cdots + 10^{n-2}+ 10^{n-1}\\ \end{array}\\\\ \begin{array}{rcl} \\ S_n &=& ( n - 0 )\cdot 10^0 +( n - 1 )\cdot 10^1 +( n - 2 )\cdot 10^2 +( n - 3 )\cdot 10^3 +( n - 4 )\cdot 10^4\\ & + & \cdots +[ n - (n-2) ]\cdot 10^{n-2} +[ n - (n-1) ]\cdot 10^{n-1}\\ S_n &=& n\cdot 1 + n\cdot 10^1 + n \cdot 10^2 + n\cdot 10^4 + \cdots + n \cdot 10^{n-2} + n \cdot 10^{n-1} \\ & -& 0\cdot 10^0 - 1 \cdot 10^1 - 2\cdot 10^2 - 3\cdot 10^3 - 4\cdot 10^4 -\cdots -(n-2)\cdot 10^{n-2} -(n-1)\cdot 10^{n-1}\\ S_n &=& n\cdot \underbrace{( 1 + 10^1 + 10^2 + 10^4 + \cdots + 10^{n-2} + 10^{n-1} )}_{\text{geometric series}} - \underbrace{\sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} }}_{\text{'Arithmetic-geometric' series}}\\ S_n &=& n\cdot \left( \dfrac{10^{n}-1}{10-1} \right) - \sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} }\\\\ S_n &=& n\cdot \left( \dfrac{10^{n}-1}{10-1} \right) - s_n\\\\ \end{array}\\ \begin{array}{lclcl} \hline s_n = \sum \limits_{r=1}^{n} { (r-1)\cdot 10^{r-1} } &=& 0\cdot 10^0 + &1& \cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 +4\cdot 10^4 +\cdots + (n-1)\cdot 10^{n-1} \\ 10\cdot s_n &=& &0&\cdot 10^1+ 1\cdot 10^2 + 2\cdot 10^3 + 3\cdot 10^4 +\cdots + (n-2)\cdot 10^{n-1} + (n-1)\cdot 10^{n} \\ \hline \end{array}\\ \begin{array}{rcl} s_n- 10 s_n &=& \underbrace{1\cdot 10^1 + 1\cdot 10^2+1\cdot 10^3 +\cdots + 1\cdot 10^{n-2} + 1\cdot 10^{n-1}}_{\text{geometric series}} - (n-1)\cdot 10^n \\\\ -9 s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{10-1} \right)- (n-1)\cdot 10^n \\\\ -9 s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{9} \right)- (n-1)\cdot 10^n \\\\ - s_n &=& 10\cdot \left( \dfrac{10^{n-1}-1}{9\cdot 9} \right)- \dfrac { (n-1)\cdot 10^n}{9} \\ \hline \end{array}\\ \begin{array}{rcl} S_n &=& n\cdot \left( \dfrac{10^{n}-1}{9} \right) - s_n\\\\ \end{array}\\\\ \begin{array}{rcl} \boxed{~ S_n = n\cdot \left( \dfrac{10^{n}-1}{9} \right) + 10\cdot \left( \dfrac{10^{n-1}-1}{9\cdot 9} \right)- \dfrac { (n-1)\cdot 10^n}{9} \\\\ \text{or }\quad S_n = \dfrac{1}{81}\cdot [~ 10\cdot (10^n - 1) - 9n ~]\\\\ \text{Example: } \\ \qquad S_4 = \dfrac{1}{81}\cdot [~ 10\cdot (10^4 - 1) - 9\cdot 4 ~]\\ \qquad S_4 = \dfrac{1}{81}\cdot [~ 10\cdot (10000 - 1) - 36 ~]\\ \qquad S_4 = \dfrac{1}{81}\cdot [~ 10\cdot (9999) - 36 ~]\\ \qquad S_4 = \dfrac{1}{81}\cdot [~ 99990 - 36 ~]\\ \qquad S_4 = \dfrac{1}{81}\cdot [~ 99954 ~]\\ \qquad S_4 = 1234 ~} \end{array} }\)

 

laugh

heureka  Nov 24, 2015

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