%5E5-binomial-expansion.jpeg "(2/x/2)^5 Binomial Expansion")
Binomial Expansion of (2/x/2)^5
The binomial theorem provides a formula for expanding expressions of the form (a + b)^n, where n is a non-negative integer. Let's apply this theorem to expand the expression (2/x/2)^5.
Simplifying the Expression
First, let's simplify the expression:
(2/x/2)^5 = (1/x)^5 = 1/x^5
Now, we can apply the binomial theorem to expand 1/x^5.
Applying the Binomial Theorem
The binomial theorem states:
(a + b)^n = ∑ (n choose k) * a^(n-k) * b^k
where:
- n choose k = n! / (k! * (n-k)!) represents the binomial coefficient
- k ranges from 0 to n
In our case, a = 1 and b = -x (since 1/x^5 = 1 + (-x)^5). Therefore, our expansion becomes:
(1 - x)^5 = ∑ (5 choose k) * 1^(5-k) * (-x)^k
Let's expand this for each value of k:
- k = 0: (5 choose 0) * 1^5 * (-x)^0 = 1
- k = 1: (5 choose 1) * 1^4 * (-x)^1 = -5x
- k = 2: (5 choose 2) * 1^3 * (-x)^2 = 10x^2
- k = 3: (5 choose 3) * 1^2 * (-x)^3 = -10x^3
- k = 4: (5 choose 4) * 1^1 * (-x)^4 = 5x^4
- k = 5: (5 choose 5) * 1^0 * (-x)^5 = -x^5
Final Expansion
Putting it all together, the expanded form of (2/x/2)^5 is:
(2/x/2)^5 = 1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5
This is the complete binomial expansion of the given expression.