91Ó°ÊÓ

The Binomial Theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Where \( \binom{n}{k} \) is the binomial coefficient. For our binomial \( \left(\frac{1}{x} + 3y\right)^5\), we identify \(a = \frac{1}{x}\), \(b = 3y\), and \(n = 5\).

Calculate the binomial coefficients \( \binom{5}{k} \) for \( k = 0, 1, 2, 3, 4, 5 \). These coefficients are: \( \binom{5}{0} = 1 \), \( \binom{5}{1} = 5 \), \( \binom{5}{2} = 10 \), \( \binom{5}{3} = 10 \), \( \binom{5}{4} = 5 \), \( \binom{5}{5} = 1 \).

Use the formula \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \) to expand the expression: \[ \left(\frac{1}{x} + 3y\right)^{5} = \sum_{k=0}^{5} \binom{5}{k} \left(\frac{1}{x}\right)^{5-k} (3y)^k \]Calculate each term:- For \(k=0\): \( \binom{5}{0} \left(\frac{1}{x}\right)^{5} (3y)^0 = \frac{1}{x^5} \)- For \(k=1\): \( \binom{5}{1} \left(\frac{1}{x}\right)^{4} (3y)^1 = 5 \left(\frac{1}{x^4}\right) 3y = \frac{15y}{x^4} \)- For \(k=2\): \( \binom{5}{2} \left(\frac{1}{x}\right)^{3} (3y)^2 = 10 \left(\frac{1}{x^3}\right) (9y^2) = \frac{90y^2}{x^3} \)- For \(k=3\): \( \binom{5}{3} \left(\frac{1}{x}\right)^{2} (3y)^3 = 10 \left(\frac{1}{x^2}\right) (27y^3) = \frac{270y^3}{x^2} \)- For \(k=4\): \( \binom{5}{4} \left(\frac{1}{x}\right)^{1} (3y)^4 = 5 \left(\frac{1}{x}\right) (81y^4) = \frac{405y^4}{x} \)- For \(k=5\): \( \binom{5}{5} \left(\frac{1}{x}\right)^{0} (3y)^5 = 1 (243y^5) = 243y^5 \)

Combine all the calculated terms:\[ \left(\frac{1}{x} + 3y\right)^{5} = \frac{1}{x^5} + \frac{15y}{x^4} + \frac{90y^2}{x^3} + \frac{270y^3}{x^2} + \frac{405y^4}{x} + 243y^5 \]