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Find the Sum of the series 1 + 2 + 9 + 64 + 625 + 7776 … till N terms

Given a number N, the task is to find the sum of the below series till N terms.
 

1 + 2 + 9 + 64 + 625 + 7776 ...

Examples: 
 

Input: N = 2 
Output:
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 1 + 2 + 9 + 64 + 625 + 7776 ... = N^{N-1}

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>


Output: 

701

 

Time Complexity: O(N * log N)

Auxiliary Space: O(1), since no extra space has been taken.

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