91Ó°ÊÓ

Recall that Pascal's Triangle is a triangular array of numbers in which each number is the sum of the two numbers directly above it. The first few rows are as follows: ``` 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 ``` From this triangle, we can find the coefficients for the expansion of \((x+y)^n\). For our problem, we need the 5th row, which is `1 5 10 10 5 1`. Now we will apply the binomial theorem, which states that for any positive integer n and any real numbers x and y, \((x+y)^n\) can be written as: \((x+y)^n = \binom{n}{0}x^n y^0 + \binom{n}{1}x^{n-1} y^1 + \cdots + \binom{n}{n}x^0 y^n\) Using the 5th row of Pascal's triangle as the coefficients, we can write the expansion of \((\mathrm{x}+2 \mathrm{y})^{5}\).

Using the coefficients from the 5th row of Pascal's Triangle and the binomial theorem, we can expand \((x+2y)^5\) as follows: \((x+2y)^5 = \binom{5}{0}x^5 y^0 + \binom{5}{1}x^4 y^1 + \binom{5}{2}x^3 y^2 + \binom{5}{3}x^2 y^3 + \binom{5}{4}x^1 y^4 + \binom{5}{5}x^0 y^5\) Now we will substitute the coefficients from the 5th row of Pascal's triangle and simplify the expression.

Substituting the coefficients and simplifying the expression, we get: \((x+2y)^5 = 1x^5(1) + 5x^4(2y) + 10x^3(2y)^2 + 10x^2(2y)^3 + 5x(2y)^4 + 1(2y)^5\) Further simplifying, we get: \((x+2y)^5 = x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5\) This is the final expanded form of \((\mathrm{x}+2 \mathrm{y})^{5}\)