(x/3+1/x)^5 Binomial Expansion

3 min read Jun 17, 2024

Expanding the Binomial (x/3 + 1/x)^5

The binomial theorem provides a formula to expand expressions of the form (a + b)^n. We can use this theorem to expand the expression (x/3 + 1/x)^5.

The binomial theorem states:

(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 choose k) = n! / (k! (n-k)!)

Let's apply the binomial theorem to our expression:

(x/3 + 1/x)^5 = ∑_(k=0)^5 (5 choose k) (x/3)^(5-k) (1/x)^k

Now, let's expand the terms for each value of k:

k = 0: (5 choose 0) (x/3)^5 (1/x)^0 = 1 * x^5 / 243 = x^5 / 243
k = 1: (5 choose 1) (x/3)^4 (1/x)^1 = 5 * x^4 / 81 * 1/x = 5x^3 / 81
k = 2: (5 choose 2) (x/3)^3 (1/x)^2 = 10 * x^3 / 27 * 1/x^2 = 10x / 27
k = 3: (5 choose 3) (x/3)^2 (1/x)^3 = 10 * x^2 / 9 * 1/x^3 = 10 / (9x)
k = 4: (5 choose 4) (x/3)^1 (1/x)^4 = 5 * x/3 * 1/x^4 = 5 / (3x^3)
k = 5: (5 choose 5) (x/3)^0 (1/x)^5 = 1 * 1 * 1/x^5 = 1/x^5

Combining all the terms, the complete expansion of (x/3 + 1/x)^5 is:

(x/3 + 1/x)^5 = x^5/243 + 5x^3/81 + 10x/27 + 10/(9x) + 5/(3x^3) + 1/x^5

This expansion can be useful for various mathematical operations and applications.