The Binomial Expansion of (x+1)^5
The binomial theorem provides a formula for expanding expressions of the form (x+y)^n, where n is a non-negative integer. In this case, we will explore the expansion of (x+1)^5.
Understanding the Binomial Theorem
The binomial theorem states:
(x + y)^n = ∑_(k=0)^n (n choose k) * x^(n-k) * y^k
where:
- (n choose k) represents the binomial coefficient, calculated as n! / (k! * (n-k)!). This represents the number of ways to choose k items from a set of n items.
- ∑_(k=0)^n indicates the sum from k=0 to n.
Applying the Theorem to (x+1)^5
Let's apply the binomial theorem to our specific case, (x+1)^5:
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Identify n: In this case, n = 5.
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Expand the sum: (x + 1)^5 = (5 choose 0) * x^5 * 1^0 + (5 choose 1) * x^4 * 1^1 + (5 choose 2) * x^3 * 1^2 + (5 choose 3) * x^2 * 1^3 + (5 choose 4) * x^1 * 1^4 + (5 choose 5) * x^0 * 1^5
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Calculate the binomial coefficients:
- (5 choose 0) = 5! / (0! * 5!) = 1
- (5 choose 1) = 5! / (1! * 4!) = 5
- (5 choose 2) = 5! / (2! * 3!) = 10
- (5 choose 3) = 5! / (3! * 2!) = 10
- (5 choose 4) = 5! / (4! * 1!) = 5
- (5 choose 5) = 5! / (5! * 0!) = 1
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Substitute the coefficients and simplify: (x + 1)^5 = 1 * x^5 * 1 + 5 * x^4 * 1 + 10 * x^3 * 1 + 10 * x^2 * 1 + 5 * x * 1 + 1 * 1 (x + 1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1
Conclusion
The binomial expansion of (x+1)^5 is x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1. This formula is useful for various mathematical applications, such as solving polynomial equations, finding derivatives, and understanding probability distributions.