## Expanding (x+4)^5: A Journey Through the Binomial Theorem

The expression (x+4)^5 might look intimidating at first glance, but fear not! We can conquer this seemingly complex expansion with the help of the **Binomial Theorem**.

### Understanding the Binomial Theorem

The Binomial Theorem provides a powerful formula to expand any expression of the form (x + y)^n, where n is a non-negative integer. The theorem states:

**(x + y)^n = ∑_(k=0)^n (n choose k) * x^(n-k) * y^k**

Let's break down this formula:

**(n choose k)**, also written as nCk or*n_C_k*, represents the binomial coefficient, calculated as n! / (k! * (n-k)!). This tells us how many ways we can choose k items from a set of n items.**∑_(k=0)^n**indicates we sum over all values of k from 0 to n.**x^(n-k) * y^k**represents the terms of the expansion, with the powers of x decreasing and the powers of y increasing.

### Applying the Theorem to (x+4)^5

Now, let's apply the Binomial Theorem to our expression (x+4)^5:

**Identify n:**In our case, n = 5.**Expand the sum:**We'll need to sum the terms for k = 0, 1, 2, 3, 4, and 5.

Let's calculate each term:

**k = 0:**(5 choose 0) * x^(5-0) * 4^0 = 1 * x^5 * 1 = x^5**k = 1:**(5 choose 1) * x^(5-1) * 4^1 = 5 * x^4 * 4 = 20x^4**k = 2:**(5 choose 2) * x^(5-2) * 4^2 = 10 * x^3 * 16 = 160x^3**k = 3:**(5 choose 3) * x^(5-3) * 4^3 = 10 * x^2 * 64 = 640x^2**k = 4:**(5 choose 4) * x^(5-4) * 4^4 = 5 * x^1 * 256 = 1280x**k = 5:**(5 choose 5) * x^(5-5) * 4^5 = 1 * x^0 * 1024 = 1024

Finally, combining all the terms, we get the expanded form of (x+4)^5:

**(x+4)^5 = x^5 + 20x^4 + 160x^3 + 640x^2 + 1280x + 1024**

### Conclusion

The Binomial Theorem provides a systematic and efficient method for expanding expressions of the form (x + y)^n. By understanding and applying this theorem, we can conquer seemingly complex expressions like (x+4)^5 and gain a deeper understanding of polynomial expansions.