Question

QUESTION 3 (6 points) (a) Use the binomial theorem to determine the coefficient of term $x^6$ in the expansion of $\left(3\sqrt[5]{x^3} - 2\right)^{12}$. (b) Use the multinomial theorem to find the constant term in the expansion of $\left(2x^2y^3 + \frac{3}{x^3} - \frac{2}{y^2} + 1\right)^{19}$.

          QUESTION 3 (6 points)
(a) Use the binomial theorem to determine the coefficient of term $x^6$ in the
expansion of
$\left(3\sqrt[5]{x^3} - 2\right)^{12}$.
(b) Use the multinomial theorem to find the constant term in the expansion of
$\left(2x^2y^3 + \frac{3}{x^3} - \frac{2}{y^2} + 1\right)^{19}$.

QUESTION 3 (6 points)
(a) Use the binomial theorem to determine the coefficient of term x^6 in the
expansion of
(3√(x^3) - 2)^12.
(b) Use the multinomial theorem to find the constant term in the expansion of
(2x^2y^3 + (3)/(x^3) - (2)/(y^2) + 1)^19.

Added by Matthew W.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Daniel Carr

05/10/2022

Step 1

We can rewrite the expression as $(3x^{3/5} - 2)^{12}$. By the binomial theorem, the general term in the expansion is given by: $$ \binom{12}{k} (3x^{3/5})^k (-2)^{12-k} = \binom{12}{k} 3^k (-2)^{12-k} x^{3k/5} $$ We want the term with $x^6$, so we need Show more…

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QUESTION 3 (6 points) (a) Use the binomial theorem to determine the coefficient of term $x^6$ in the expansion of $(3\sqrt[5]{x^3} - 2)^{12}$. (b) Use the multinomial theorem to find the constant term in the expansion of $(2x^2y^3 + \frac{3}{x^3} - \frac{2}{y^2} + 1)^{19}$.