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suzi
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How do you find the 3rd term of the expansion of (2x-y)^5
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The formula for finding the 3rd term of the expansion of (2x-y)^5 is (5 choose 2)(2x)^3(-y)^2, which can be simplified to 120x^3y^2.
To use the binomial theorem, you would first write out the expansion of (2x-y)^5, which is (5 choose 0)(2x)^5(-y)^0 + (5 choose 1)(2x)^4(-y)^1 + (5 choose 2)(2x)^3(-y)^2 + (5 choose 3)(2x)^2(-y)^3 + (5 choose 4)(2x)^1(-y)^4 + (5 choose 5)(2x)^0(-y)^5. Then, you would simply take the 3rd term, which is (5 choose 2)(2x)^3(-y)^2, and simplify it to get 120x^3y^2.
(5 choose 2) represents the number of ways you can choose 2 elements from a set of 5 elements, also known as a combination. In this case, it is used to determine the coefficient of the 3rd term in the expansion of (2x-y)^5.
No, it is not necessary to use the binomial theorem to find the 3rd term. You can also use the formula (n choose k)(a)^k(b)^(n-k), where n is the exponent, k is the term you want to find, a is the coefficient of the first term, and b is the coefficient of the second term. In this case, it would be (5 choose 2)(2)^3(-1)^2, which simplifies to 120x^3y^2.
Yes, the same method can be used to find the 3rd term of the expansion of any binomial raised to a power. You would simply plug in the appropriate values for n, k, a, and b in the formula (n choose k)(a)^k(b)^(n-k).