# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

(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 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 = 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; ` `    ``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 = 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``;  ` `    ``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

Output:

```55
```

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