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Problem 732

Expand $$(\mathrm{x}+2 \mathrm{y})^{5}$$

Expert verified
$$(x+2y)^5 = x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5$$
See the step by step solution

Step 1: Recall Pascal's Triangle and Binomial theorem

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 $$(\mathrm{x}+2 \mathrm{y})^{5}$$.

Step 2: Write the expansion using the coefficients and the binomial theorem

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.

Step 3: Substitute coefficients and simplify

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 $$(\mathrm{x}+2 \mathrm{y})^{5}$$

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