The given expression is $(a + b)^5$, which can be expanded using the Binomial Theorem. The theorem states:
$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k,$ where $\binom{n}{k}$ represents the binomial coefficient.

Expansion of $(a + b)^5$:

For $n = 5$, the expansion is: $(a + b)^5 = \binom{5}{0}a^5b^0 + \binom{5}{1}a^4b^1 + \binom{5}{2}a^3b^2 + \binom{5}{3}a^2b^3 + \binom{5}{4}a^1b^4 + \binom{5}{5}a^0b^5.$

Simplifying coefficients: $= 1 \cdot a^5 + 5 \cdot a^4b + 10 \cdot a^3b^2 + 10 \cdot a^2b^3 + 5 \cdot ab^4 + 1 \cdot b^5.$

Final Result:

$(a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5.$

Would you like additional details about this expansion?

Related Questions:

1. How can the binomial coefficient $\binom{n}{k}$ be calculated?
2. What is the general formula for $(a + b)^n$ for higher powers $n$?
3. How can the expansion of $(a + b)^5$ be visualized in Pascal's Triangle?
4. How would this expansion differ if $a = 1$ and $b = 1$?
5. How does the binomial theorem apply to negative exponents?

Tip:

Use Pascal’s Triangle for quick computation of binomial coefficients for small values of $n$.