Final answer:
The binomial expansion of (x + (1)/(x))^5 is x^5 + 5x^3 + 5x + 5/3x + 5(1/x) + (1/x)^5.
Step-by-step explanation:
The binomial expansion of (x + \frac{1}{x})^5 can be found using the binomial theorem. The general formula for the binomial expansion of (a + b)^n is:
(a + b)^n = C(n, 0)a^n b^0 + C(n, 1)a^(n-1) b^1 + C(n, 2)a^(n-2) b^2 + ... + C(n, n-1)a^1 b^(n-1) + C(n, n)a^0 b^n
Where C(n, k) represents the binomial coefficient and is calculated using the formula:
C(n, k) = \frac{n!}{k!(n-k)!}
For our given expression, a = x, b = \frac{1}{x}, and n = 5. We can substitute these values into the formula to find each term in its simplest form.
Step-by-step solution:
- Calculate the binomial coefficient for each term:
- C(5, 0) = \frac{5!}{0!(5-0)!} = \frac{1}{1} = 1
- C(5, 1) = \frac{5!}{1!(5-1)!} = \frac{5}{1} = 5
- C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{10}{2} = 5
- C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{10}{6} = \frac{5}{3}
- C(5, 4) = \frac{5!}{4!(5-4)!} = \frac{5}{1} = 5
- C(5, 5) = \frac{5!}{5!(5-5)!} = \frac{1}{1} = 1
Write out each term:
- Term 1: 1 * x^5 * \left(\frac{1}{x}\right)^0 = x^5
- Term 2: 5 * x^4 * \left(\frac{1}{x}\right)^1 = 5x^3
- Term 3: 5 * x^3 * \left(\frac{1}{x}\right)^2 = 5x
- Term 4: \frac{5}{3} * x^2 * \left(\frac{1}{x}\right)^3 = \frac{5}{3x}
- Term 5: 5 * x^1 * \left(\frac{1}{x}\right)^4 = 5\left(\frac{1}{x}\right)
- Term 6: 1 * x^0 * \left(\frac{1}{x}\right)^5 = \left(\frac{1}{x}\right)^5
Combine the terms:
(x + \frac{1}{x})^5 = x^5 + 5x^3 + 5x + \frac{5}{3x} + 5\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^5