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Sum of kth powers of first n natural numbers

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  • Last Updated : 21 Nov, 2022
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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++




// 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

55

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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