The numerically greatest term in the expansio | vedclass

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Question

(A) Given the expansion $(3x - 4y)^{23}$. Substituting $x = \frac{1}{6}$ and $y = \frac{1}{8}$,we get $(3(\frac{1}{6}) - 4(\frac{1}{8}))^{23} = (\frac

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$^{23}C_{11} \cdot (\frac{1}{2})^{23}$

Vedclass · Answer

(A) Given the expansion $(3x - 4y)^{23}$. Substituting $x = \frac{1}{6}$ and $y = \frac{1}{8}$,we get $(3(\frac{1}{6}) - 4(\frac{1}{8}))^{23} = (\frac{1}{2} - \frac{1}{2})^{23} = 0^{23}$.However,for the numerically greatest term in $(a+b)^n$,we consider the absolute values $|T_{r+1}| = |^{n}C_r a^{n-r} b^r|$.Let $a = 3x = \frac{1}{2}$ and $b = -4y = -\frac{1}{2}$.$|T_{r+1}| = |^{23}C_r (\frac{1}{2})^{23-r} (-\frac{1}{2})^r| = ^{23}C_r (\frac{1}{2})^{23}$.To find the greatest term,we maximize $^{23}C_r$. For $n=23$,the maximum value of $^{23}C_r$ occurs at $r = \frac{n}{2} \pm 0.5$,i.e.,$r = 11$ and $r = 12$.Thus,the greatest terms are $T_{12}$ and $T_{13}$,both equal to $^{23}C_{11} (\frac{1}{2})^{23}$.

The numerically greatest term in the expansion of $(3x - 4y)^{23}$ when $x = \frac{1}{6}$ and $y = \frac{1}{8}$ is:

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