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++ 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 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 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 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# 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 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 ?> |
<script> // Javascript program to find the sum of // the series 1+11+111+1111+.... // Function for finding summation
function summation( n) {
let sum = 0, j = 1;
for ( let i = 1; i <= n; i++) {
sum = sum + j;
// Appending a 1 at the end
j = (j * 10) + 1;
}
return sum;
}
// Driver Code
let n = 5;
document.write(summation(n));
// This code contributed by Princi Singh </script> |
Output :
12345
Time Complexity: O(n), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Another method: Let given a series S = 1 + 11 + 111 + 1111 + . . . + upto nth term. Using formula to find sum of series.
Below is the implementation of above approach.
// 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 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 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# 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 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 ?> |
<script> // javascript program to find the sum of // the series 1+11+111+1111+.... // Function for finding summation function summation( n)
{ let sum;
sum = (Math.pow(10, n + 1) -
10 - (9 * n)) / 81;
return sum;
} // Driver Code let n = 5; document.write(summation(n)) ;
// This code is contributed by aashish1995 </script> |
Output :
12345
Time Complexity: O(logn), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.