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Exploring the Sum of Fourth Powers of the First \(n\) Odd Natural Numbers

In the realm of number theory and mathematical series, understanding the sum of powers of natural numbers is a fundamental concept. While we are familiar with sums like the sum of the first (n) natural numbers ((\sum_{k = 1}^{n}k=\frac{n(n + 1)}{2})) or the sum of squares ((\sum_{k=1}^{n}k^{2}=\frac{n(n + 1)(2n + 1)}{6})), the sum of fourth powers of odd natural numbers has its own unique charm and applications. In this blog, we will delve deep into deriving the formula for the sum of fourth powers of the first (n) odd natural numbers, understand its derivation steps, and see some example usages.

2026-07

Table of Content#

  1. Representation of Odd Natural Numbers
  2. Derivation of the Formula
    • Using the binomial expansion approach
    • Subtracting the sum of fourth powers of even natural numbers
  3. Simplification of the Formula
  4. Example Usage
  5. Best Practices and Common Pitfalls
  6. References

1. Representation of Odd Natural Numbers#

The (k^{th}) odd natural number can be represented as (a_{k}=2k - 1), where (k = 1,2,\cdots,n). So, the sum of fourth powers of the first (n) odd natural numbers is (S=\sum_{k = 1}^{n}(2k-1)^{4})

2. Derivation of the Formula#

Using the binomial expansion approach#

Expanding ((2k-1)^{4}) using the binomial theorem:

[ (2k-1)^{4}=16k^{4}-32k^{3}+24k^{2}-8k + 1 ]

Therefore:

[ \sum_{k = 1}^{n}(2k-1)^{4}=16\sum_{k=1}^{n}k^{4}-32\sum_{k=1}^{n}k^{3}+24\sum_{k=1}^{n}k^{2}-8\sum_{k=1}^{n}k+\sum_{k=1}^{n}1 ]

We know the standard sum formulas:

[ \sum_{k=1}^{n}k=\frac{n(n + 1)}{2},\quad \sum_{k=1}^{n}k^{2}=\frac{n(n + 1)(2n + 1)}{6},\quad \sum_{k=1}^{n}k^{3}=\left(\frac{n(n + 1)}{2}\right)^{2},\quad \sum_{k=1}^{n}k^{4}=\frac{n(n + 1)(2n + 1)(3n^{2}+3n-1)}{30} ]

Substituting these into the expansion:

[ \begin{align*} \sum_{k=1}^{n}(2k-1)^{4}&=16\times\frac{n(n + 1)(2n + 1)(3n^{2}+3n-1)}{30}-32\times\left(\frac{n(n + 1)}{2}\right)^{2}+24\times\frac{n(n + 1)(2n + 1)}{6}-8\times\frac{n(n + 1)}{2}+n \end{align*} ]

Subtracting the sum of fourth powers of even natural numbers#

Alternatively, we can express the sum in terms of the sum of all natural numbers minus the even numbers. The sum of fourth powers of the first (2n) natural numbers is:

[ \sum_{k=1}^{2n}k^{4}=\frac{2n(2n + 1)(4n+1)(12n^{2}+6n - 1)}{30} ]

The sum of fourth powers of even natural numbers (the first (n) even numbers) is:

[ \sum_{k = 1}^{n}(2k)^{4}=16\sum_{k=1}^{n}k^{4}=16\times\frac{n(n + 1)(2n + 1)(3n^{2}+3n - 1)}{30} ]

Since (\sum_{k = 1}^{2n}k^{4}=\sum_{k = 1}^{n}(2k-1)^{4}+\sum_{k = 1}^{n}(2k)^{4}), we have:

[ \sum_{k = 1}^{n}(2k-1)^{4}=\sum_{k = 1}^{2n}k^{4}-\sum_{k = 1}^{n}(2k)^{4} ]

[ \begin{align*} S&=\frac{2n(2n + 1)(4n + 1)(12n^{2}+6n-1)}{30}-16\times\frac{n(n + 1)(2n + 1)(3n^{2}+3n - 1)}{30}\ &=\frac{n(2n+1)}{30}\left[(2)(4n + 1)(12n^{2}+6n-1)-16(n + 1)(3n^{2}+3n-1)\right] \end{align*} ]

3. Simplification of the Formula#

After expanding and simplifying the above expression, we obtain the final formula:

[ \begin{align*} \sum_{k=1}^{n}(2k-1)^{4}&=\frac{n(2n-1)(2n+1)(12n^{2}-7)}{15} \end{align*} ]

We can also expand it in polynomial form:

[ \begin{align*} S&=\frac{n(2n-1)(2n+1)(12n^{2}-7)}{15}\ &=\frac{24n^{5}-10n^{3}+n}{15} \end{align*} ]

4. Example Usage#

Example 1: Let (n = 1)

Using the formula (S=\frac{n(2n-1)(2n+1)(12n^{2}-7)}{15}):

[ \begin{align*} S&=\frac{1\times(2\times1-1)(2\times1 + 1)(12\times1^{2}-7)}{15}\ &=\frac{1\times1\times3\times5}{15}\ &= \frac{15}{15}=1 \end{align*} ]

And ((2\times1 - 1)^{4}=1^{4}=1). The formula gives the correct result.

Example 2: Let (n=2)

[ \begin{align*} S&=\frac{2\times(2\times2-1)(2\times2 + 1)(12\times2^{2}-7)}{15}\ &=\frac{2\times3\times5\times41}{15}\ &=\frac{246}{15}=82 \end{align*} ]

And ((2\times1-1)^{4}+(2\times2 - 1)^{4}=1^{4}+3^{4}=1 + 81=82). The formula is correct.

Example 3: Let (n=3)

[ \begin{align*} S&=\frac{3\times(2\times3-1)(2\times3 + 1)(12\times3^{2}-7)}{15}\ &=\frac{3\times5\times7\times113}{15}\ &=\frac{11865}{15}=707 \end{align*} ]

And ((2\times1-1)^{4}+(2\times2-1)^{4}+(2\times3-1)^{4}=1^{4}+3^{4}+5^{4}=1+81+625=707).

5. Best Practices and Common Pitfalls#

  • Best Practices:
    • Understand the Basics: Have a clear understanding of the sum formulas for powers of natural numbers ((\sum_{k = 1}^{n}k^{m})) before attempting to derive the sum of powers of odd or even numbers.
    • Check with Small Values: Always check the derived formula with small values of (n) (e.g., (n=1,2,3)) to ensure its correctness.
  • Common Pitfalls:
    • Formula Substitution Errors: When subtracting or adding series, make sure to substitute the correct limits and formulas. For example, when subtracting the sum of even-numbered series from the sum of all series, ensure that the number of terms is consistent.
    • Arithmetic Mistakes: During the expansion and simplification of polynomial expressions (which is common in series formula derivations), be careful with arithmetic operations like multiplication, addition, and subtraction.

6. References#

  • Book: "Concrete Mathematics: A Foundation for Computer Science" by Ronald L. Graham, Donald E. Knuth, and Oren Patashnik. This book covers various techniques for dealing with sums and recurrences.
  • Online Resources: MathWorld (https://mathworld.wolfram.com/) has detailed pages on sums of powers of natural numbers and related concepts.