Program to find value of 1^k + 2^k + 3^k + … + n^k

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Given two positive integers N and K. The task is to evaluate the value of 1^K + 2 ^K + 3^K + … + N^K.

Examples:

Input: N = 3, K = 4
Output: 98
Explanation:
?(x⁴) = 1⁴ + 2⁴ + 3⁴, where 1 ? x ? N
?(x⁴) = 1 + 16 + 81
?(x⁴) = 98

Input: N = 8, K = 4
Output: 8772

Approach:

The idea is to find the value of x^K using pow() function. (where x is from 1 to N).
The required sum is the summation of all values calculated above.

Below is the implementation of the above approach:

C++

// C++ Program to find the value
// 1^K + 2^K + 3^K + .. + N^K
#include <bits/stdc++.h>
using namespace std;
 
// Function to find value of
// 1^K + 2^K + 3^K + .. + N^K
int findSum(int N, int k)
{
    // Initialise sum to 0
    int sum = 0;
    for (int i = 1; i <= N; i++) {
 
        // Find the value of
        // pow(i, 4) and then
        // add it to the sum
        sum += pow(i, k);
    }
 
    // Return the sum
    return sum;
}
 
// Drivers Code
int main()
{
    int N = 8, k = 4;
 
    // Function call to
    // find the sum
    cout << findSum(N, k) << endl;
    return 0;
}

Java

// Java Program to find the value
// 1^K + 2^K + 3^K + .. + N^K
class GFG {
     
    // Function to find value of
    // 1^K + 2^K + 3^K + .. + N^K
    static int findSum(int N, int k)
    {
        // Initialise sum to 0
        int sum = 0;
        for (int i = 1; i <= N; i++) {
     
            // Find the value of
            // pow(i, 4) and then
            // add it to the sum
            sum += (int)Math.pow(i, k);
        }
     
        // Return the sum
        return sum;
    }
     
    // Drivers Code
    public static void main (String[] args) 
    {
        int N = 8, k = 4;
     
        // Function call to
        // find the sum
        System.out.println(findSum(N, k));
    }
}
 
// This code is contributed by AnkitRai01

Python 3

# Python 3 Program to find the value
# 1^K + 2^K + 3^K + .. + N^K
from math import pow
 
# Function to find value of
# 1^K + 2^K + 3^K + .. + N^K
def findSum(N, k):
     
    # Initialise sum to 0
    sum = 0
    for i in range(1, N + 1, 1):
         
        # Find the value of
        # pow(i, 4) and then
        # add it to the sum
        sum += pow(i, k)
 
    # Return the sum
    return sum
 
# Drives Code
if __name__ == '__main__':
    N = 8
    k = 4
 
    # Function call to
    # find the sum
    print(int(findSum(N, k)))
 
# This code is contributed by Surendra_Gangwar

C#

// C# Program to find the value
// 1^K + 2^K + 3^K + .. + N^K
 
using System;
 
public class GFG {
     
    // Function to find value of
    // 1^K + 2^K + 3^K + .. + N^K
    static int findSum(int N, int k)
    {
        // Initialise sum to 0
        int sum = 0;
        for (int i = 1; i <= N; i++) {
     
            // Find the value of
            // pow(i, 4) and then
            // add it to the sum
            sum += (int)Math.Pow(i, k);
        }
     
        // Return the sum
        return sum;
    }
     
    // Drivers Code
    public static void Main (string[] args) 
    {
        int N = 8, k = 4;
     
        // Function call to
        // find the sum
        Console.WriteLine(findSum(N, k));
    }
 
}
 
// This code is contributed by AnkitRai01

Javascript

<script>
    // Javascript Program to find the value
    // 1^K + 2^K + 3^K + .. + N^K
     
    // Function to find value of
    // 1^K + 2^K + 3^K + .. + N^K
    function findSum(N, k)
    {
        // Initialise sum to 0
        let sum = 0;
        for (let i = 1; i <= N; i++) {
       
            // Find the value of
            // pow(i, 4) and then
            // add it to the sum
            sum += Math.pow(i, k);
        }
       
        // Return the sum
        return sum;
    }
     
    let N = 8, k = 4;
       
    // Function call to
    // find the sum
    document.write(findSum(N, k));
 
// This code is contributed by divyesh072019.
</script>

Output:

Time Complexity: O(N * log(k)) (here we iterate for loop from i = 1 to N and we need to calculate pow (i,k) which required log(k) time so overall time complexity will be O(N * log(k)) )
Auxiliary space: O(1)

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