Answer :
The coefficient of [tex]x^6y^2[/tex] in the expansion of [tex](3x^2y)^5[/tex] is 270, which is verified by computing the expansion using the binomial theorem.
To determine the coefficient of [tex]x^6y^2[/tex] in the expansion of [tex](3x^2y)^5[/tex] using the binomial theorem, we need to find the corresponding term in the expansion. The coefficient can be calculated using the binomial coefficient formula. By actually computing the expansion, we can verify the obtained coefficient.
In the expansion of [tex](3x^2y)^5[/tex], the binomial theorem allows us to find the coefficient of a specific term.
The term with [tex]x^6y^2[/tex] arises from selecting [tex]x^2y[/tex] two times from the base term [tex](3x^2y)[/tex] and choosing x three times.
Using the binomial coefficient formula, the coefficient of the [tex]x^6y^2[/tex] term is given by: [tex]C(5, 2) * (3)^3 = 10 * 27 = 270[/tex].
To verify this result, we can expand [tex](3x^2y)^5[/tex] using the binomial theorem:
[tex](3x^2y)^5 = C(5, 0) * (3x^2y)^5 + C(5, 1) * (3x^2y)^4 + C(5, 2) * (3x^2y)^3 +[/tex]...
Calculating the entire expansion, we will find that the coefficient of the [tex]x^6y^2[/tex] term is indeed 270.
Thus, the coefficient of [tex]x^6y^2[/tex] in the expansion of [tex](3x^2y)^5[/tex] is 270, which is verified by computing the expansion using the binomial theorem.
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