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Question

Find the numerically greatest term in the expansion of $$\left ( 3-5x \right )^{15}$$ when $$\displaystyle x=\frac{1}{5}.$$


Solution

Let $$r^{th}$$ and $$\left ( r+1 \right )^{th}$$ be two consecutive terms in the expansion of $$\left ( 3-5x \right )^{15}$$
$$T_{r+1}\geq T_{r}$$
$$^{15}\textrm{C}_{r}3^{15-r}\left ( \left | -5x \right | \right )^{r}\geq ^{15}\textrm{C}_{r-1}3^{15-\left ( r-1 \right )}\left ( \left | -5x \right | \right )^{r-1}$$
$$\displaystyle \frac{\left ( 15 \right )!}{\left ( 15-r \right )!r!}\left | -5x \right |\geq \frac{3.\left ( 15 \right )!}{\left ( 16-r \right )!\left ( r-1 \right )!}$$
$$\displaystyle 5.\frac{1}{5}\left ( 16-r \right )\geq 3r$$
$$16-r\geq 3r$$
$$4r\leq 16$$
$$r\leq 4$$
Hence, $$4$$th term is the greatest.

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