Expand \((x + 3y)^6\) using binomial expansion, then evaluate \((1.03)^6\).

To expand \((x + 3y)^6\) using binomial expansion, we can use the formula:

\[(x + y)^n = \binom{n}{0}x^n y^0 + \binom{n}{1}x^{n-1} y^1 + \binom{n}{2}x^{n-2} y^2 + \ldots + \binom{n}{n}x^0 y^n\]

In this case, \(n = 6\), \(x = x\), and \(y = 3y\). So, the expansion of \((x + 3y)^6\) is:

\[\binom{6}{0}x^6 (3y)^0 + \binom{6}{1}x^5 (3y)^1 + \binom{6}{2}x^4 (3y)^2 + \binom{6}{3}x^3 (3y)^3 + \binom{6}{4}x^2 (3y)^4 + \binom{6}{5}x (3y)^5 + \binom{6}{6} (3y)^6\]

Simplifying each term, we get:

\[x^6 + 18x^5y + 135x^4y^2 + 540x^3y^3 + 1215x^2y^4 + 1458xy^5 + 729y^6\]

Now, to evaluate \((1.03)^6\), we can substitute \(x = 1\) and \(y = 0.03\) into the expanded expression:

\[1^6 + 18(1)^5(0.03) + 135(1)^4(0.03)^2 + 540(1)^3(0.03)^3 + 1215(1)^2(0.03)^4 + 1458(1)(0.03)^5 + (0.03)^6\]

Simplifying each term, we get:

\[1 + 0.54 + 1.215 + 1.215 + 1.215 + 0.4374 + 0.000729\]

Adding all these terms together, we get:

\[1 + 0.54 + 1.215 + 1.215 + 1.215 + 0.4374 + 0.000729 = 5.623129\]

Therefore, \((1.03)^6 \approx 5.623129\).

Using the binomial expansion, we have:

$$\begin{split} (x 3y)^6 & = \sum_{k=0}^6 \binom{6}{k} x^{6-k} (3y)^k \\\ & = \binom{6}{0} x^6 + \binom{6}{1} x^5 3y + \binom{6}{2} x^4 (3y)^2 + \binom{6}{3} x^3 (3y)^3 \\\ & \quad + \binom{6}{4} x^2 (3y)^4 + \binom{6}{5} x (3y)^5 + \binom{6}{6} (3y)^6 \\\ & = x^6 + 18x^5y + 162x^4y^2 + 666x^3y^3 + 1620x^2y^4 + 2430xy^5 + 2187y^6. \end{split}$$

To evaluate \((1.03)^6\), we simply substitute \(x = 1.03\) and \(y = 0\) into the expansion:

$$(1.03)^6 = 1.03^6 + 18(1.03)^5(0) + 162(1.03)^4(0)^2 + 666(1.03)^3(0)^3 $$

$$+ 1620(1.03)^2(0)^4 + 2430(1.03)(0)^5 + 2187(0)^6 = 1.194052.$$