前n个自然数的k次幂之和

给定两个整数n和k ，任务是计算并打印1 ^k + 2 ^k + 3 ^k +…+ n ^k 。
例子：

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 
 

方法：证明前n个自然数的平方和：

(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 
 

添加所有方程式：

(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 
 

同样，可以通过以下方式显示多维数据集的证明：

(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 
 

同样，对于第k次幂和，

(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 
 

下面是上述方法的实现：

C++

// C++ program to find sum pf k-th powers of
// first n natural numbers.
#include 
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 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..

Python3

# 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

C#

// 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
}

Javascript

 
// Javascript program to find sum pf 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
 

输出：

55