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Find the sum of the series 1+11+111+1111+….. upto n terms
  • Last Updated : 24 Apr, 2020

Here we are going to find the sum of the series 1 + 11 + 111 + 1111 +…..upto N terms (where N is given).

Example :

Input : 3
Output : 1 + 11 + 111 +....
Total sum is : 123

Input : 4
Output : 1 + 11 + 111 + 1111 +..... 
Total sum is : 1234

Input : 7
Output : 1 + 11 + 111 + 1111 + 11111 + 
         111111 + 1111111 +..... 
Total sum is : 1234567


Here we see that when value of N is 3, series last upto 1 + 11 + 111 i.e, three term and it’s sum is 123.

Program for finding sum of above series :

C++






// C++ program to find the sum of
// the series 1+11+111+1111+....
#include <bits/stdc++.h>
using namespace std;
  
// Function for finding summation
int summation(int n)
{
    int sum = 0, j = 1;
    for (int i = 1; i <= n; i++) {
        sum = sum + j;
  
        // Appending a 1 at the end
        j = (j * 10) + 1;
    }
  
    return sum;
}
  
// Driver Code
int main()
{
    int n = 5;
    cout << " " <<  summation(n);
    return 0;
}
  
// This code is contributed by shivanisinghss2110


C




// C program to find the sum of
// the series 1+11+111+1111+....
#include <stdio.h>
  
// Function for finding summation
int summation(int n)
{
    int sum = 0, j = 1;
    for (int i = 1; i <= n; i++) {
        sum = sum + j;
  
        // Appending a 1 at the end
        j = (j * 10) + 1;
    }
  
    return sum;
}
  
// Driver Code
int main()
{
    int n = 5;
    printf("%d", summation(n));
    return 0;
}


Java




// Java program to find the sum of
// the series 1+11+111+1111+....
import java.io.*;
  
class GFG 
{
  
    // Function for finding summation
    static int summation(int n)
    {
        int sum = 0, j = 1;
        for (int i = 1; i <= n; i++) 
        {
            sum = sum + j;
            j = (j * 10) + 1;
        }
  
        return sum;
    }
  
    // Driver Code
    public static void main(String args[])
    {
        int n = 5;
        System.out.println(summation(n));
    }
}
  
// This code is contributed
// by Nikita Tiwari


Python




# Python program to get the summation
# of following series
def summation(n):
    sum = 0
    j = 1
      
    for i in range(1, n + 1):
        sum = sum + j
        j = (j * 10) + 1
          
    return sum
          
# Driver Code
n = 5
print(summation(n))


C#




// C# program to find the sum of
// the series 1+11+111+1111+....
using System;
  
class GFG 
{
  
    // Function for finding summation
    static int summation(int n)
    {
        int sum = 0, j = 1;
        for (int i = 1; i <= n; i++) 
        {
            sum = sum + j;
            j = (j * 10) + 1;
        }
  
        return sum;
    }
  
    // Driver Code
    public static void Main()
    {
        int n = 5;
        Console.WriteLine(summation(n));
    }
}
  
// This code is contributed by vt_m


PHP




<?php
// PHP program to find the sum of
// the series 1+11+111+1111+....
  
// Function for finding summation
function summation($n)
{
    $sum = 0; $j = 1;
    for ($i = 1; $i <= $n; $i++)
    {
        $sum = $sum + $j;
  
        // Appending a 1 at the end
        $j = ($j * 10) + 1;
    }
  
    return $sum;
}
  
// Driver Code
$n = 5;
echo summation($n);
  
// This code is contributed by ajit
?>



Output :

12345

Another method: Let given a series S = 1 + 11 + 111 + 1111 + . . . + upto nth term. Using formula to find sum of series.

  \\ S = 1 + 11 + 111 + 1111 + . . . + upto \ n \ term \\ \\  S = \left ( \frac{1}{9} \right )*\left ( 9 + 99 + 999 + 9999 + . . . + upto \ n \ term \right ) \\ \\ S = \left ( \frac{1}{9} \right ) * \left ( \left (10 ^{1} - 1\right ) + \left (10 ^{2} -1 \right ) +\left (10 ^{3} -1 \right ) + \left (10 ^{4} -1 \right ) + . . . + \left (10 ^{n} -1 \right ) \right ) \\ \\ S = \left ( \frac{1}{9} \right ) *\left (10 ^{1} + 10 ^{2} + 10 ^{3} + 10 ^{4} + . . . + 10 ^{n} - n\right ) \\ \\ S = \left ( \frac{1}{9} \right ) * \left ( \frac{10 * (10^{n} - 1)}{10 - 1} - n \right ) \\ \\ S = \frac{10^{n+1} - 10 - 9n}{81}

Below is the implementation of above approach.

C++




// C++ program to find the sum of
// the series 1+11+111+1111+....
#include <bits/stdc++.h>
  
// Function for finding summation
int summation(int n)
{
    int sum;
  
    sum = (pow(10, n + 1) - 
               10 - (9 * n)) / 81;
  
    return sum;
}
  
// Driver Code
int main()
{
    int n = 5;
    printf("%d", summation(n));
    return 0;
}


Java




// java program to find the sum of
// the series 1+11+111+1111+....
import java.io.*;
  
class GFG {
  
    // Function for finding summation
    static int summation(int n)
    {
        int sum;
      
        sum = (int)(Math.pow(10, n + 1) - 
                10 - (9 * n)) / 81;
      
        return sum;
    }
      
    // Driver Code
    public static void main (String[] args)
    {
        int n = 5;
        System.out.println(summation(n));
    }
}
  
// This code is contributed by anuj_67.


Python3




# Python3 program to 
# find the sum of
# the series 1+11+111+1111+....
import math
  
# Function for 
# finding summation
def summation(n):
    return int((pow(10, n + 1) - 
                    10 - (9 * n)) / 81);
  
# Driver Code
print(summation(5));
  
# This code is contributed
# by mits.


C#




// C# program to find the sum of
// the series 1+11+111+1111+....
using System;
  
class GFG {
  
    // Function for finding summation
    static int summation(int n)
    {
        int sum;
      
        sum = (int)(Math.Pow(10, n + 1) - 
                10 - (9 * n)) / 81;
      
        return sum;
    }
      
    // Driver Code
    public static void Main ()
    {
        int n = 5;
        Console.WriteLine(summation(n));
    }
}
  
// This code is contributed by anuj_67.


PHP




<?php
//PHP program to find the sum of
// the series 1+11+111+1111+....
  
// Function for finding summation
function summation($n)
{
    $sum;
  
    $sum = (pow(10, $n + 1) - 
                10 - (9 * $n)) / 81;
  
    return $sum;
}
  
// Driver Code
$n = 5;
echo summation($n);
  
// This code is contributed by aj_36
?>



Output :

12345

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