Sum of Fourth Powers of first N natural numbers

The fourth power of a number x is x raised to the power 4 or x4. Natural numbers are all positive integers excluding zero. Thus, the sum of the fourth powers of the first N natural numbers is −

$\mathrm{Sum = 1^4 + 2^4 + 3^4 + 4^4 + … + N^4}$

This article describes some approaches for finding the sum using minimum time and space complexity.

Problem Statement

Given the number N, find the sum $\mathrm{1^4 + 2^4 + 3^4 + 4^4 + … + N^4}$.

Example 1

Input: 3

Output: 98

Explanation

$\mathrm{Sum = 1^4 + 2^4 + 3^4 = 1 + 16 + 81 = 98}$

Example 2

Input: 5

Output: 979

Explanation

$\mathrm{Sum = 1^4 + 2^4 + 3^4 + 4^4 + 5^4 = 1 + 16 + 81 + 256 + 625 = 979}$

Solution 1: Brute Force Approach

The brute force approach for solving the problem will be to simply calculate the fourth powers of all numbers from 1 to N and then add them.

Pseudocode

procedure fourthSum (n)
   sum = 0
   for i = 1 to n:
      sum = sum + i^4
   end for
end procedure

Example

In the following program, we’ll calculate the value of x raised to power 4 from x ranging from 1 to n and then add them.

#include<bits/stdc++.h>
using namespace std;
// Function to find the sum of fourth powers of first n natural numbers
long long fourthSum(long long n){

   // initializing the sum variable
   long long sum = 0;
   
   // adding the fourth powers of all numbers from 1 to n to sum
   for (int i = 1 ; i <= n ; i++) {
   
      // calculating i raised to the power 4
      // pow() function returns double thus converting it to int
      sum += int(pow(i,4));
   }
   return sum;
}
int main(){
   long long N = 7;
   cout << "Sum of fourth powers of first " << N << " natural numbers = ";
   cout << fourthSum(N);
}

Output

Sum of fourth powers of first 7 natural numbers = 4676

Time Complexity − O(n) as the pow() function is executed for n times with time complexity of O(4) i.e. constant each. Thus n*O(4) will be O(n).

Space Complexity − O(1) as no extra space is used.

Solution 2: Faulhaber’s Formula

Faulhaber’s Formula gives the general formula for the power sum of the first n positive integers.

$$\mathrm{\displaystyle\sum\limits_{k=1}^n k^p=1^p+2^p+3^p+...+n^p}$$

Deriving the formula for$\mathrm{\sum_{k=1}^nk^4}$:

We know that, $\mathrm{\sum_{k=1}^n k=\frac{n(n+1)}{2},\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6},\sum_{k=1}^nk^3=\frac{n(n+1)^2}{4}}$

Now, using binomial expansion

$\mathrm{(𝑛 + 1)^5 − 𝑛^5 = 5𝑛^4 + 10𝑛^3 + 10𝑛^2 + 5n + 1….(1)}$

Similarly,

$\mathrm{n^5 −(n − 1)^5 = 5(n − 1)^4 + 10(n − 1)\ 3 + 10(n − 1)^2 + 5(n − 1) + 1 …..(2)}$

Computing till we will have n equations

$\mathrm{2^5 − 1^5 = 5 \cdot 1^4 + 10 \cdot 1^3 + 10 \cdot 1^2 + 5 \cdot 1 + 1 ….(n)}$

Adding both LHS and RHS of equations 1 to n −

$$\mathrm{(n+1)^5−1\cdot\displaystyle\sum\limits_{k=1}^n k^4+10\cdot\displaystyle\sum\limits_{k=1}^n k^3+10\cdot\displaystyle\sum\limits_{k=1}^n k^2+5\cdot\displaystyle\sum\limits_{k=1}^n k+n}$$

Replacing the values of $\mathrm{\sum_{k=1}^nk,\sum_{k=1}^nk^2,\:and\:\sum_{k=1}^nk^3,}$ in the above equation, we can find $\mathrm{\sum_{k=1}^nk^4}$

$$\mathrm{\displaystyle\sum\limits_{k=1}^n k^4=\frac{(n+1)^5−1−10\cdot\sum^n_{k=1}k^3−10\cdot\sum^n_{k=1}k^2−5\cdot\sum^n_{k=1}k−n}{5}}$$

Thus,

$$\mathrm{\displaystyle\sum\limits_{k=1}^n k^4=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}−\frac{n}{30}=\frac{6\cdot n^5+15\cdot n^4+10\cdot n^3−n}{30}}$$

We’ll be using the above formula to calculate the sum of the fourth powers of the first N natural numbers.

Example 1

Input: 3

Output: 98

Explanation

n⁵ = 243, n⁴ = 81, n³ = 27, n = 3. Thus, sum = ((6*243) + (15*81) + (10*27) - 3)/30 = 98

Example 2

Input: 9

Output: 15333

Explanation

n⁵ = 59049, n⁴ = 6561, n³ = 729, n = 9. Thus, sum = ((6*59049) + (15*6561) + (10*729) - 9)/30 = 15333

Pseudocode −

procedure fourthSum (n)
   sum = 0
   fifth power = n5
   fourth power = n4
   third power = n3
   sum = ((6 * fifth power) + (15 * fourth power) + (10 * third power) - n)/30
end procedure

Example

In the following program, we’ll calculate the sum using Faulhaber’s Formula that is derived above.

#include<bits/stdc++.h>
using namespace std;

// Function to find the sum of fourth powers of first n natural numbers
long long fourthSum(long long n){

   // initializing the sum variable
   long long sum = 0;
   
   // calculating n raised to the power 5
   // pow() returns double thus changing it to int
   long long fifthPower = int(pow(n,5));
   
   // calculating n raised to the power 4
   long long fourthPower = int(pow(n,4));
   
   // calculating n raised to the power 3
   long long thirdPower = int(pow(n,3));
   sum = ((6 * fifthPower) + (15 * fourthPower) + (10 * thirdPower) - n)/30;
   return sum;
}

int main(){
   long long N = 11;
   cout << "Sum of fourth powers of first " << N << " natural numbers = ";
   cout << fourthSum(N);
}

Output

Sum of fourth powers of first 11 natural numbers = 39974

Time Complexity − O(1) as we are using the direct formula for calculating the sum.

Space Complexity − O(1) as no extra space is used.

Conclusion

In conclusion, in order to find the sum of the fourth powers of the first N natural numbers, we can follow two approaches. The first is to simply calculate the powers and add them. But this approach takes a linear time complexity. Another way to solve it using constant time is using Faulhaber’s formula.