Sum of kth powers of first n natural numbers

0 2 10 minutes read

Given two integers n and k, the task is to calculate and print 1^k + 2^k + 3^k + … + n^k.
Examples:

Input: n = 5, k = 2
Output: 55
1² + 2² + 3² + 4² + 5² = 1 + 4 + 9 + 16 + 25 = 55
Input: n = 10, k = 4
Output: 25333

Approach: Proof for sum of squares of first n natural numbers:

(n+1)³ – n³ = 3 * (n²) + 3 * n + 1
putting n = 1, 2, 3, 4, …, n
2³ – 1³ = 3 * (1²) + 3 * 1 + 1 …equation 1
3³ – 2³ = 3 * (2²) + 3 * 2 + 1 …equation 2
4³ – 3³ = 3 * (3²) + 3 * 3 + 1 …equation 3
……
……
……
(n + 1)³ – n³ = 3 * (n²) + 3 * n + 1 …equation n

Adding all equations:

(n + 1)³ – 1³ = 3 * (sum of square terms) + 3 * (sum of n terms) + n
n³ + 3 * n² + 3 * n = 3 * (sum of square terms) + (3 * n * (n + 1)) / 2 + n
sum of square terms = (n * (n + 1) * (2 * n + 1)) / 6

Similarly, proof for cubes can be shown by taking:

(n+1)⁴ – n⁴ = 4 * n³ + 6 * n² + 4 * n + 1
if we continue this process for n⁵, n⁶, n⁷ … n^k
Sum of squares,
(n + 1)³ – 1³ = ³C₁ * sum(n²) + ³C₂ * sum(n) + ³C₃
Using sum of squares we can find the sum of cubes,
(n + 1)⁴ – 1⁴ = ⁴C₁ * sum(n³) + ⁴C₂ * sum(n²) + ⁴C₃ * sum(n) + ⁴C₄

Similarly for kth powers sum,

(n + 1)^k – 1^k = ^kC₁ * sum(n^{(k – 1)}) + ^kC₂ * sum(n^{(k – 2)}) + … + ^kC_{(k – 1)} * (sum(n^(k-(k-1)) + ^kC_k * n
where C stands for binomial coefficients
Use modulus function for higher values of n

Below is the implementation of the above approach:

C++

// C++ program to find sum of k-th powers of 
// first n natural numbers.
#include <bits/stdc++.h>
using namespace std;
 
// A global array to store factorials
const int MAX_K = 15;
long long unsigned int fac[MAX_K];
 
// Function to calculate the factorials
// of all the numbers upto k
void factorial(int k)
{
    fac[0] = 1;
    for (int i = 1; i <= k + 1; i++) {
        fac[i] = (i * fac[i - 1]);
    }
}
 
// Function to return the binomial coefficient
long long unsigned int bin(int a, int b)
{
 
    // nCr = (n! * (n - r)!) / r!
    long long unsigned int ans =
               ((fac[a]) / (fac[a - b] * fac[b]));
    return ans;
}
 
// Function to return the sum of kth powers of 
// n natural numbers
long int sumofn(int n, int k)
{
    int p = 0;
    long long unsigned int num1, temp, arr[1000];
    for (int j = 1; j <= k; j++) {
 
        // When j is unity
        if (j == 1) {
            num1 = (n * (n + 1)) / 2;
 
            // Calculating sum(n^1) of unity powers
            // of n; storing sum(n^1) for sum(n^2)
            arr[p++] = num1;
 
            // If k = 1 then temp is the result
            temp = num1;
        }
        else {
            temp = (pow(n + 1, j + 1) - 1 - n);
 
            // For finding sum(n^k) removing 1 and
            // n * kCk from (n + 1)^k
            for (int s = 1; s < j; s++) {
 
                // Removing all kC2 * sum(n^(k - 2))
                // + ... + kCk - 1 * (sum(n^(k - (k - 1))
                temp = temp -
                    (arr[j - s - 1] * bin(j + 1, s + 1));
            }
            temp = temp / (j + 1);
 
            // Storing the result for next sum of
            // next powers of k
            arr[p++] = temp;
        }
    }
    temp = arr[p - 1];
    return temp;
}
 
// Driver code
int main()
{
    int n = 5, k = 2;
    factorial(k);
    cout << sumofn(n, k) << "\n";
    return 0;
}

Java

// Java program to find sum of k-th powers of 
// first n natural numbers.
 
import java.io.*;
 
class GFG {
 
 
// A global array to store factorials
static int MAX_K = 15;
static int fac[] = new int[MAX_K];
 
// Function to calculate the factorials
// of all the numbers upto k
static void factorial(int k)
{
    fac[0] = 1;
    for (int i = 1; i <= k + 1; i++) {
        fac[i] = (i * fac[i - 1]);
    }
}
 
// Function to return the binomial coefficient
static  int bin(int a, int b)
{
 
    // nCr = (n! * (n - r)!) / r!
    int ans =
            ((fac[a]) / (fac[a - b] * fac[b]));
    return ans;
}
 
// Function to return the sum of kth powers of 
// n natural numbers
static int sumofn(int n, int k)
{
    int p = 0;
    int num1, temp;
    int arr[] = new int[1000];
    for (int j = 1; j <= k; j++) {
 
        // When j is unity
        if (j == 1) {
            num1 = (n * (n + 1)) / 2;
 
            // Calculating sum(n^1) of unity powers
            // of n; storing sum(n^1) for sum(n^2)
            arr[p++] = num1;
 
            // If k = 1 then temp is the result
            temp = num1;
        }
        else {
            temp = ((int)Math.pow(n + 1, j + 1) - 1 - n);
 
            // For finding sum(n^k) removing 1 and
            // n * kCk from (n + 1)^k
            for (int s = 1; s < j; s++) {
 
                // Removing all kC2 * sum(n^(k - 2))
                // + ... + kCk - 1 * (sum(n^(k - (k - 1))
                temp = temp -
                    (arr[j - s - 1] * bin(j + 1, s + 1));
            }
            temp = temp / (j + 1);
 
            // Storing the result for next sum of
            // next powers of k
            arr[p++] = temp;
        }
    }
    temp = arr[p - 1];
    return temp;
}
 
// Driver code
 
    public static void main (String[] args) {
            int n = 5, k = 2;
    factorial(k);
    System.out.println( sumofn(n, k));
    }
}
// This code is contributed by anuj_67..

Python3

# Python3 program to find the sum of k-th
# powers of first n natural numbers
 
# global array to store factorials 
MAX_K = 15
fac = [1 for i in range(MAX_K)]
 
# function to calculate the factorials 
# of all the numbers upto k
def factorial(k):
    fac[0] = 1
    for i in range(1, k + 2):
        fac[i] = (i * fac[i - 1])
 
# function to return the binomial coeff
def bin(a, b):
     
    # nCr=(n!*(n-r)!)/r!
    ans = fac[a] // (fac[a - b] * fac[b])
    return ans
     
# function to return the sum of the kth 
# powers of n natural numbers
def sumofn(n, k):
    p = 0
    num1, temp = 1, 1
    arr = [1 for i in range(1000)]
     
    for j in range(1, k + 1):
         
        # when j is 1
        if j == 1:
            num1 = (n * (n + 1)) // 2
             
            # calculating sum(n^1) of unity powers
            #of n storing sum(n^1) for sum(n^2)
            arr[p] = num1
            p += 1
             
            # if k==1 then temp is the result
        else:
            temp = pow(n + 1, j + 1) - 1 - n
             
            # for finding sum(n^k) removing 1 and
            # n*KCk from (n+1)^k
            for s in range(1, j):
                 
                # Removing all kC2 * sum(n^(k - 2))
                # + ... + kCk - 1 * (sum(n^(k - (k - 1))
                temp = temp - (arr[j - s - 1] *
                               bin(j + 1, s + 1))
            temp = temp // (j + 1)
             
            # storing the result for next sum 
            # of next powers of k
            arr[p] = temp
            p += 1
    temp = arr[p - 1]
    return temp
 
# Driver code
n, k = 5, 2
factorial(k)
print(sumofn(n, k))
 
# This code is contributed by Mohit kumar 29 

C#

// C# program to find sum of k-th powers of 
// first n natural numbers. 
 
using System;
 
class GFG { 
 
// A global array to store factorials 
static int MAX_K = 15; 
static int []fac = new int[MAX_K]; 
 
// Function to calculate the factorials 
// of all the numbers upto k 
static void factorial(int k) 
{ 
    fac[0] = 1; 
    for (int i = 1; i <= k + 1; i++) { 
        fac[i] = (i * fac[i - 1]); 
    } 
} 
 
// Function to return the binomial coefficient 
static int bin(int a, int b) 
{ 
 
    // nCr = (n! * (n - r)!) / r! 
    int ans = 
            ((fac[a]) / (fac[a - b] * fac[b])); 
    return ans; 
} 
 
// Function to return the sum of kth powers of 
// n natural numbers 
static int sumofn(int n, int k) 
{ 
    int p = 0; 
    int num1, temp; 
    int []arr = new int[1000]; 
    for (int j = 1; j <= k; j++) { 
 
        // When j is unity 
        if (j == 1) { 
            num1 = (n * (n + 1)) / 2; 
 
            // Calculating sum(n^1) of unity powers 
            // of n; storing sum(n^1) for sum(n^2) 
            arr[p++] = num1; 
 
            // If k = 1 then temp is the result 
            temp = num1; 
        } 
        else { 
            temp = ((int)Math.Pow(n + 1, j + 1) - 1 - n); 
 
            // For finding sum(n^k) removing 1 and 
            // n * kCk from (n + 1)^k 
            for (int s = 1; s < j; s++) { 
 
                // Removing all kC2 * sum(n^(k - 2)) 
                // + ... + kCk - 1 * (sum(n^(k - (k - 1)) 
                temp = temp - 
                    (arr[j - s - 1] * bin(j + 1, s + 1)); 
            } 
            temp = temp / (j + 1); 
 
            // Storing the result for next sum of 
            // next powers of k 
            arr[p++] = temp; 
        } 
    } 
    temp = arr[p - 1]; 
    return temp; 
} 
 
    // Driver code 
    public static void Main () { 
            int n = 5, k = 2; 
            factorial(k); 
            Console.WriteLine(sumofn(n, k)); 
    } 
    // This code is contributed by Ryuga 
} 

PHP

<?php
// PHP program to find sum of k-th powers 
// of first n natural numbers.
 
// A global array to store factorial
$MAX_K = 15;
$fac[$MAX_K] = array();
 
// Function to calculate the factorials
// of all the numbers upto k
function factorial($k)
{
    global $fac;
    $fac[0] = 1;
    for ($i = 1; $i <= $k + 1; $i++) 
    {
        $fac[$i] = ($i * $fac[$i - 1]);
    }
}
 
// Function to return the binomial 
// coefficient
function bin($a, $b)
{
    global $MAX_K;
    global $fac;
     
    // nCr = (n! * (n - r)!) / r!
    $ans = (($fac[$a]) / ($fac[$a - $b] *
                          $fac[$b]));
    return $ans;
}
 
// Function to return the sum of kth 
// powers of n natural numbers
function sumofn($n, $k)
{
    $p = 0; $num1; $temp;
    $arr[1000] = array();
    for ($j = 1; $j <= $k; $j++) 
    {
 
        // When j is unity
        if ($j == 1) 
        {
            $num1 = ($n * ($n + 1)) / 2;
 
            // Calculating sum(n^1) of unity powers
            // of n; storing sum(n^1) for sum(n^2)
            $arr[$p++] = $num1;
 
            // If k = 1 then temp is the result
            $temp = $num1;
        }
        else
        {
            $temp = (pow($n + 1, $j + 1) - 1 - $n);
 
            // For finding sum(n^k) removing 1 
            // and n * kCk from (n + 1)^k
            for ($s = 1; $s < $j; $s++)
            {
 
                // Removing all kC2 * sum(n^(k - 2))
                // + ... + kCk - 1 * (sum(n^(k - (k - 1))
                $temp = $temp - ($arr[$j - $s - 1] * 
                                  bin($j + 1, $s + 1));
            }
            $temp = $temp / ($j + 1);
 
            // Storing the result for next 
            // sum of next powers of k
            $arr[$p++] = $temp;
        }
    }
    $temp = $arr[$p - 1];
    return $temp;
}
 
// Driver code
$n = 5;
$k = 2;
factorial($k);
echo sumofn($n, $k), "\n";
 
// This code is contributed by Sachin
?>

Javascript

<script>
 
// Javascript program to find sum of k-th powers of 
// first n natural numbers.    
 
// A global array to store factorials
    var MAX_K = 15;
    var fac = Array(MAX_K).fill(0);
 
    // Function to calculate the factorials
    // of all the numbers upto k
    function factorial(k) {
        fac[0] = 1;
        for (i = 1; i <= k + 1; i++) {
            fac[i] = (i * fac[i - 1]);
        }
    }
 
    // Function to return the binomial coefficient
    function bin(a , b) {
 
        // nCr = (n! * (n - r)!) / r!
        var ans = ((fac[a]) / (fac[a - b] * fac[b]));
        return ans;
    }
 
    // Function to return the sum of kth powers of
    // n natural numbers
    function sumofn(n , k) {
        var p = 0;
        var num1, temp;
        var arr = Array(1000).fill(0);
        for (j = 1; j <= k; j++) {
 
            // When j is unity
            if (j == 1) {
                num1 = (n * (n + 1)) / 2;
 
                // Calculating sum(n^1) of unity powers
                // of n; storing sum(n^1) for sum(n^2)
                arr[p++] = num1;
 
                // If k = 1 then temp is the result
                temp = num1;
            } else {
                temp = (parseInt( Math.pow(n + 1, j + 1) - 1 - n));
 
                // For finding sum(n^k) removing 1 and
                // n * kCk from (n + 1)^k
                for (s = 1; s < j; s++) {
 
                    // Removing all kC2 * sum(n^(k - 2))
                    // + ... + kCk - 1 * (sum(n^(k - (k - 1))
                    temp = temp - (arr[j - s - 1] * bin(j + 1, s + 1));
                }
                temp = temp / (j + 1);
 
                // Storing the result for next sum of
                // next powers of k
                arr[p++] = temp;
            }
        }
        temp = arr[p - 1];
        return temp;
    }
 
    // Driver code
 
     
        var n = 5, k = 2;
        factorial(k);
        document.write(sumofn(n, k));
 
// This code contributed by Rajput-Ji
 
</script>

Output

Complexity Analysis:

Time Complexity: O(k*(log(k)+k)).

In the worst case as we can see in sumofn() function we have a log component and a loop nested so time complexity would be O(k*(log(k)+k)) in the worst-case.

Space Complexity: We have declared a global array of size 15 and an array of size 1000 is declared inside sumofn() function, so space complexity is O(p) where p=(15+1000).

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