Explanation:
A The given expansion of the binomial (3\(5 \mathrm{x})^{11}\)
Now, the expression can be written as-
\(3^{11}\left(1-\frac{5 x}{3}\right)^{11}\)
We have the expansion of the binomial \(\left(1-\frac{5 x}{3}\right)^{11}\)
Then, \(\quad \frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}}=\frac{12-\mathrm{r}}{\mathrm{r}}\left|\frac{-5 \mathrm{x}}{3}\right|\)
When, \(\mathrm{x}=\frac{1}{5}\)
Then,
\(\frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}}=\frac{12-\mathrm{r}}{\mathrm{r}}\left(\frac{1}{3}\right)\)
\(\frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}}=\frac{12-\mathrm{r}}{3 \mathrm{r}} \geq 1\)
\(\frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}} \geq 1\)
\(\frac{12-\mathrm{r}}{3 \mathrm{r}} \geq 1\)
\(4 \mathrm{r} \leq 12\)
\(\mathrm{r} \leq 3\)
From this relation \(r=2,3\)
So, the greatest terms are \(\mathrm{T}_{2+1}\) and \(\mathrm{T}_{3+1}\)
And, \(\mathrm{T}_{3+1}\) is the greatest term-
\(\therefore \therefore \quad 3^{11}\left(\left|\mathrm{~T}_{3+1}\right|\right)\)
\(=3^{11}\left(\left.{ }^{11} \mathrm{C}_3\left(\frac{-5}{3} \mathrm{x}\right)^3 \right\rvert\,\right.\)
\(=3^{11}\left|{ }^{11} \mathrm{C}_3\left(\frac{-5}{3} \mathrm{x}\right)^3\right|\)
\(=3^{11}\left|{ }^{11} \mathrm{C}_3\left(\frac{-5}{3} \times \frac{1}{5}\right)^3\right| \quad\left(\because \mathrm{x}=\frac{1}{5}(\text { Given })\right)\)
\(\left.=3^{11} \frac{11 \times 10 \times 9}{1 \times 2 \times 3}\left(\frac{-1}{27}\right) \right\rvert\,=55 \times 3^9\)