Given a number N. The task is to find the sum of the below series up to N-th term:
1 + 22 + 333 + 4444 + …up to n terms
Examples:
Input: N = 3 Output: 356 Input: N = 10 Output: 12208504795
Approach:
Below is the implementation of the above approach:
C++
// CPP program to find the sum // of given series#include <iostream>#include <math.h>using namespace std;// Function to calculate sumint findSum(int n){ // Return sum return (pow(10, n + 1) * (9 * n - 1) + 10) / pow(9, 3) - n * (n + 1) / 18;}// Driver codeint main(){ int n = 3; cout << findSum(n); return 0;} |
Java
// Java Program to find // Sum of first n termsimport java.util.*;class solution{static int calculateSum(int n){// Returning the final sumreturn ((int)Math.pow(10, n + 1) * (9 * n - 1) + 10) / (int)Math.pow(9, 3) - n * (n + 1) / 18;}// Driver codepublic static void main(String ar[]){// no. of terms to find the sumint n=3;System.out.println("Sum= "+ calculateSum(n));}}//This code is contributed by Surendra_Gangwar |
Python 3
# Python program to find the sum of given series.# Function to calculate sumdef solve_sum(n): # Return sum return (pow(10, n + 1)*(9 * n - 1)+10)/pow(9, 3)-n*(n + 1)/18# driver coden = 3print(int(solve_sum(n))) |
C#
// C# Program to find // Sum of first n termsusing System;class solution{static int calculateSum(int n){// Returning the final sumreturn ((int)Math.Pow(10, n + 1) * (9 * n - 1) + 10) / (int)Math.Pow(9, 3) - n * (n + 1) / 18;}// Driver codepublic static void Main(){// no. of terms to find the sumint n=3;Console.WriteLine("Sum= "+ calculateSum(n));}}//This code is contributed by inder_verma. |
PHP
<?php// PHP program to find the sum // of given series// Function to calculate sumfunction findSum($n){ // Return sum return (pow(10, $n + 1) * (9 * $n - 1) + 10) / pow(9, 3) - $n * ($n + 1) / 18;}// Driver code$n = 3;echo findSum($n);// This code is contributed // by inder_verma.?> |
Javascript
<script>// Javascript Program to find // Sum of first n terms function calculateSum( n) { // Returning the const sum return (parseInt(Math.pow(10, n + 1)) * (9 * n - 1) + 10) / parseInt(Math.pow(9, 3)) - n * (n + 1) / 18; } // Driver code // no. of terms to find the sum let n = 3; document.write("Sum= " + calculateSum(n));// This code is contributed by 29AjayKumar </script> |
Output:
356
Time Complexity: O(logn), where n represents the given integer, as we have used pow function.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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