%5E99.jpeg "(a+b)^99")
Exploring the Expansion of (a + b)^99
The expression (a + b)^99 may seem intimidating at first glance, but with the help of the Binomial Theorem, we can systematically expand it into a sum of terms. This theorem provides a powerful tool for understanding the expansion of any binomial raised to a positive integer power.
The Binomial Theorem
The Binomial Theorem states that for any real numbers a and b, and any non-negative integer n:
(a + b)^n = ∑_(k=0)^n (n choose k) a^(n-k) b^k
where (n choose k) is the binomial coefficient, calculated as n!/(k!(n-k)!). This coefficient represents the number of ways to choose k elements from a set of n elements.
Applying the Theorem to (a + b)^99
To expand (a + b)^99 using the Binomial Theorem, we simply need to substitute n with 99 and apply the formula:
(a + b)^99 = ∑_(k=0)^99 (99 choose k) a^(99-k) b^k
This expands into a sum of 100 terms, each with a specific coefficient and powers of a and b.
Key Observations
- Symmetry: Notice the symmetry in the binomial coefficients. For example, (99 choose 0) = (99 choose 99), (99 choose 1) = (99 choose 98), and so on. This implies that the expansion exhibits a mirrored structure.
- Patterns: The powers of a decrease from 99 to 0, while the powers of b increase from 0 to 99.
- Combinatorial Interpretation: Each term in the expansion corresponds to a unique way of choosing k bs from a total of 99 factors in the original expression.
Example: First Few Terms
Let's look at the first few terms of the expansion:
(99 choose 0) a^99 b^0 + (99 choose 1) a^98 b^1 + (99 choose 2) a^97 b^2 + ...
This simplifies to:
a^99 + 99a^98b + 4851a^97b^2 + ...
Conclusion
The Binomial Theorem provides a powerful and elegant way to understand the expansion of (a + b)^99 and similar expressions. It unveils the underlying structure and patterns, enabling us to efficiently calculate and analyze the resulting terms. While the full expansion might be lengthy, the theorem allows us to focus on specific terms or patterns based on our needs.