Given a number N, the task is to find the sum of the below series till N terms.
Examples:
Input: N = 2
Output: 3
1 + 2 = 3
Input: N = 5
Output: 701
1 + 2 + 9 + 64 + 625 = 701
Approach: From the given series, find the formula for Nth term:
1st term = 1 = 11-1 2nd term = 2 = 22-1 3rd term = 9 = 33-1 4th term = 64 = 44-1 . . Nth term = NN - 1
Therefore:
Nth term of the series
Then iterate over numbers in the range [1, N] to find all the terms using the above formula and compute their sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the sum of series void printSeriesSum( int N) { long long sum = 0; for ( int i = 1; i <= N; i++) { // Generate the ith term and // add it to the sum sum += pow (i, i - 1); } // Print the sum cout << sum << endl; } // Driver Code int main() { int N = 5; printSeriesSum(N); return 0; } |
Java
// Java program for the above approach class GFG{ // Function to find the sum of series static void printSeriesSum( int N) { long sum = 0 ; for ( int i = 1 ; i <= N; i++) { // Generate the ith term and // add it to the sum sum += Math.pow(i, i - 1 ); } // Print the sum System.out.print(sum + "\n" ); } // Driver Code public static void main(String[] args) { int N = 5 ; printSeriesSum(N); } } // This code is contributed by Princi Singh |
Python3
# Python3 program for the above approach # Function to find the summ of series def printSeriessumm(N): summ = 0 for i in range ( 1 ,N + 1 ): # Generate the ith term and # add it to the summ summ + = pow (i, i - 1 ) # Print the summ print (summ) # Driver Code N = 5 printSeriessumm(N) # This code is contributed by shubhamsingh10 |
C#
// C# program for the above approach using System; class GFG{ // Function to find the sum of series static void printSeriesSum( int N) { double sum = 0; for ( int i = 1; i <= N; i++) { // Generate the ith term and // add it to the sum sum += Math.Pow(i, i - 1); } // Print the sum Console.Write(sum + "\n" ); } // Driver Code public static void Main(String[] args) { int N = 5; printSeriesSum(N); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // javascript program for the above approach // Function to find the sum of series function printSeriesSum( N) { let sum = 0; for (let i = 1; i <= N; i++) { // Generate the ith term and // add it to the sum sum += Math.pow(i, i - 1); } // Print the sum document.write(sum); } // Driver Code let N = 5; printSeriesSum(N); // This code is contributed by todaysgaurav </script> |
701
Time Complexity: O(N * log N)
Auxiliary Space: O(1), since no extra space has been taken.
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