First slide

Binomial theorem for positive integral Index

Question

$The numericall6y greatest term in the expansion of (3 + 2 x)^{49} when x = 1 / 5 is$

Moderate

A

4th term
B

5th term
C

6th term
D

7th term

Solution

$\begin{array}{l} (3 + 2 x)^{49} = 3^{49} (1 + 2 x / 3)^{49} = 3^{49} (1 + α)^{49} where α = \frac{2 x}{3} = \frac{2 (1 / 5)}{3} = \frac{2}{15} \\ \frac{(n + 1) | α |}{| α | + 1} = \frac{100 / 15}{17 / 15} = \frac{100}{17} = 5 + \frac{15}{17} \\ ∴ 6^{th} term is the numerically greatest. \end{array}$

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