The greatest term in the expansion of (3 + 5x)15, when x = 1/5 is

The greatest term in the expansion of ${(3 + 5 x)}^{15}$ , when $x = 1 / 5$ is

$α = \frac{(15 + 1) (5) (1 / 5)}{3 + 5 (1 / 5)} = 4$

Thus, 4th and 5th terms are the greatest, and $T_{5} =^{15} C_{4} (3)^{11}$

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