My approach :$$T(r) = \binom{n}{r-1}(3x)^{n-r+1} (-2)^{r-1}$$here $T(r)$ is the $r$th term of the binomial expansion of $(3x-2)^{21}$.$$T(r) = \binom{n}{r-1}(3)^{n-r+1} (-2)^{r-1}(x)^{n-r+1}$$here $n = 21$ , so , $$T(r) = \binom{21}{r-1}(3)^{22-r} (-2)^{r-1}(x)^{n-r+1}$$So we need $\binom{21}{r-1}(3)^{22-r}(-2)^{r-1}$ to be maximum. I am not able to go further, please help me in this method or provide another method to solve the question in title.
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$\begingroup$Compare the term $3^a2^{21-a}\begin{pmatrix}21\\a\\\end{pmatrix}$ with $3^{a+1}2^{20-a}\begin{pmatrix}21\\a+1\\\end{pmatrix}$. The ratio is $$\frac{3(21-a)}{2(a+1)}.$$Since this is greater than $1$ for $a\le 12$ the maximum value occurs when $a=12$ i.e. $$3^{12}2^{9}\begin{pmatrix}21\\12\\\end{pmatrix}.$$
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