Calculating the Sum of the Fourth Powers of the First n Natural Numbers

Understanding how to calculate the sum of the fourth powers of the first n natural numbers is a fundamental topic in mathematics. This article will explore the derivation of the formula, provide examples, and explain the reasoning behind the calculations.

Derivation of the Formula

There are multiple ways to derive the formula for the sum of the fourth powers of the first n natural numbers. One common method is the use of mathematical induction, but another convenient approach is to use known formulas for sums of powers. Many mathematical references and textbooks also provide this formula.

The formula for the sum of the fourth powers is:

[ sum_{k1}^{n} k^4 frac{n(n 1)(2n 1)(3n^2 3n-1)}{30} ]

Example Calculation

To illustrate the formula, let's consider the case where n 5. The sum of the fourth powers for the first 5 natural numbers can be calculated as follows:

Substitute n 5 into the formula:

[ sum_{k1}^{5} k^4 frac{5(5 1)(2 cdot 5 1)(3 cdot 5^2 3 cdot 5 - 1)}{30} ]

Breaking down the formula step-by-step:

n 5 n 1 6 2n 1 11 3n2 3n-1 3 cdot 25 3 cdot 5 - 1 75 15 - 1 89

Substitute these values into the formula:

[ sum_{k1}^{5} k^4 frac{5 cdot 6 cdot 11 cdot 89}{30} ]

Calculate the numerator:

[ 5 cdot 6 cdot 11 cdot 89 30 cdot 11 cdot 89 330 cdot 89 29595 ]

Finally, divide by 30:

[ frac{29595}{30} 986.5 ]

So, the sum of the fourth powers from 1 to 5 is:

[ 1^4 2^4 3^4 4^4 5^4 1 16 81 256 625 979 ]

This matches our formula calculation.

The School Level Trick

The school level trick involves expressing each term as a product of some successive terms of an arithmetic progression (AP) and using similar identities for the sums of cubes and squares. Let's explore this method with the given identity:

[ n^4 frac{n(n 1)(n-1)(3n^2 - 3n - 1)}{5} - 2 cdot frac{n(n 1)(n-1)}{4} cdot n frac{3n(n 1)}{2} cdot n^2 ]

The formula can be derived step-by-step as follows:

Starting with the expression:

[ n^4 frac{n(n 1)(n-1)(3n^2 - 3n - 1)}{5} - 2 cdot frac{n(n 1)(n-1)}{4} cdot n frac{3n(n 1)}{2} cdot n^2 ]

By simplifying and summing the series, we get:

[ sum_{k1}^{n} k^4 frac{n(n 1)(2n 1)(3n^2 3n-1)}{30} ]

This confirms the formula using the school level trick.

Conclusion

This formula can be used to calculate the sum of the fourth powers for any positive integer n. Mastering this method can be incredibly useful for solving complex problems in mathematics and related fields.