Find the sum of the series 1+11+111+1111+….. upto n terms
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 ?> |
Javascript
<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++
// 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 ?> |
Javascript
<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.