The absolute value of the numerically greates | vedclass

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Question

(B) Given the expression $(2x - 3y)^{12}$ with $x = 3$ and $y = 2$. Substituting the values,we get $(2(3) - 3(2))^{12} = (6 - 6)^{12} = 0^{12}$. Howev

Vedclass · Accepted Answer

${}^{12}C_6 6^{12}$

Vedclass · Answer

(B) Given the expression $(2x - 3y)^{12}$ with $x = 3$ and $y = 2$. Substituting the values,we get $(2(3) - 3(2))^{12} = (6 - 6)^{12} = 0^{12}$. However,the question asks for the numerically greatest term in the expansion of $(2x - 3y)^{12}$ for specific values. Let $T_{r+1} = {}^{12}C_r (2x)^{12-r} (-3y)^r$. For $x=3, y=2$,$T_{r+1} = {}^{12}C_r (6)^{12-r} (-6)^r = {}^{12}C_r (6)^{12} (-1)^r$. The absolute value is $|T_{r+1}| = {}^{12}C_r 6^{12}$. This value is maximized when ${}^{12}C_r$ is maximum,which occurs at $r = \frac{12}{2} = 6$. Thus,the greatest term is ${}^{12}C_6 6^{12}$.

The absolute value of the numerically greatest term in the expansion of $(2x - 3y)^{12}$ when $x = 3$ and $y = 2$ is:

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