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The equation in the image shows the formula for the sum of the fourth powers of the first \( n \) natural numbers:

\[
1^4 + 2^4 + 3^4 + \dots + n^4 = \frac{n(n+1)(6n^3 + 9n^2 + n - 1)}{30}
\]

This formula gives a compact expression for the sum of the fourth powers of integers from 1 to \( n \). 

Would you like a step-by-step explanation of how this formula is derived, or do you have any specific questions about it?

Here are some related questions to explore:
1. How is the general formula for sums of powers of integers derived?
2. Can this formula be generalized to other powers (e.g., third, fifth powers)?
3. What is the closed form for the sum of cubes \(1^3 + 2^3 + \dots + n^3\)?
4. How is this formula useful in solving problems in calculus or number theory?
5. Can we verify this formula by calculating it for a small value of \(n\)?

**Tip**: Summation formulas like this often arise from polynomial fits or using methods such as the method of finite differences!

What is the formula for the sum of the fourth powers of the first n natural numbers?

Learn the formula for the sum of the fourth powers of the first n natural numbers: 1^4 + 2^4 + ... + n^4 = n(n+1)(6n^3 + 9n^2 + n - 1)/30. This topic covers key concepts in summation of powers and polynomial functions. Suitable for undergraduate mathematics students.

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