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.Maths

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