Sum of kth powers of first n natural numbers

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

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

Below is the implementation of the above approach:

C++

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// C++ program to find sum pf 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;
}

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Java

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// Java program to find sum pf 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..

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Python3

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# Python3 program to find the sum pf 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 

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C#

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// C# program to find sum pf 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 

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PHP

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<?php
// PHP program to find sum pf 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
?>

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Output:

55

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