In the expansion of {left( {3x - frac{1}{{{x^ | vedclass

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(C) To find the $r^{th}$ term from the end in the expansion of $(a + b)^n$,we can rewrite the expression as $(b + a)^n$ and find the $r^{th}$ term fro

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$\frac{17010}{x^8}$

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(C) To find the $r^{th}$ term from the end in the expansion of $(a + b)^n$,we can rewrite the expression as $(b + a)^n$ and find the $r^{th}$ term from the beginning.Here,the expression is $(3x - x^{-2})^{10}$.Reversing the terms,we get $(-x^{-2} + 3x)^{10}$.The $r^{th}$ term from the beginning is given by $T_r = {^{n}C_{r-1}} (a)^{n-(r-1)} (b)^{r-1}$.For the $5^{th}$ term from the end $(r=5)$,we use $n=10$,$a = -x^{-2}$,and $b = 3x$.$T_5 = {^{10}C_{4}} (-x^{-2})^{10-4} (3x)^4$.$T_5 = {^{10}C_{4}} (-x^{-2})^6 (3x)^4$.$T_5 = 210 	imes (x^{-12}) 	imes (81x^4)$.$T_5 = 210 	imes 81 	imes x^{-12+4}$.$T_5 = 17010 	imes x^{-8} = \frac{17010}{x^8}$.

In the expansion of ${\left( {3x - \frac{1}{{{x^2}}}} \right)^{10}}$,the $5^{th}$ term from the end is:

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