The numerically greatest term in the expansio | vedclass

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(C) Given the expansion $(2x - 3y)^{13}$ with $x = \frac{7}{2}$ and $y = \frac{3}{7}$.Substituting the values,we get $(2(\frac{7}{2}) - 3(\frac{3}{7})

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$26 \cdot 3^5 \cdot 7^9$

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(C) Given the expansion $(2x - 3y)^{13}$ with $x = \frac{7}{2}$ and $y = \frac{3}{7}$.Substituting the values,we get $(2(\frac{7}{2}) - 3(\frac{3}{7}))^{13} = (7 - \frac{9}{7})^{13} = (\frac{49-9}{7})^{13} = (\frac{40}{7})^{13}$.However,for the numerically greatest term in $(a+b)^n$,we consider the absolute values of the terms.Let $T_{r+1}$ be the $(r+1)$-th term in the expansion of $(a+b)^n$.The ratio $\left| \frac{T_{r+1}}{T_r} \right| = \frac{n-r+1}{r} \left| \frac{b}{a} \right|$.Here $a = 7$,$b = -\frac{9}{7}$,and $n = 13$.$\left| \frac{T_{r+1}}{T_r} \right| = \frac{13-r+1}{r} \left| \frac{-9/7}{7} \right| = \frac{14-r}{r} \cdot \frac{9}{49} = \frac{9(14-r)}{49r}$.For the term to be increasing,$\frac{9(14-r)}{49r} \geq 1 \implies 126 - 9r \geq 49r \implies 126 \geq 58r \implies r \leq \frac{126}{58} \approx 2.17$.Thus,$r = 2$ gives the greatest term $T_{2+1} = T_3$.$T_3 = \binom{13}{2} (7)^{11} (-\frac{9}{7})^2 = \frac{13 \cdot 12}{2} \cdot 7^{11} \cdot \frac{81}{49} = 78 \cdot 7^9 \cdot 81 = 78 \cdot 3^4 \cdot 7^9 = 26 \cdot 3 \cdot 3^4 \cdot 7^9 = 26 \cdot 3^5 \cdot 7^9$.

The numerically greatest term in the expansion of $(2x - 3y)^{13}$ when $x = \frac{7}{2}$ and $y = \frac{3}{7}$ is:

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