First slide

Binomial theorem for positive integral Index

Question

The two consecutive terms in the expansion of (3x + 2)⁷⁴, whose coefficients are equal, are

Moderate

A

20th and 21st
B

30th and 31st
C

40th and 41st
D

none of these

Solution

Since, $\frac{T_{3}}{T_{2}} = 7$ (given)
$\begin{array}{l} \Rightarrow \frac{^{n} C_{2} {(2^{x})}^{n - 2} {(4^{- x})}^{2}}{^{n} C_{1} {(2^{x})}^{n - 1} \cdot (4^{- x})} = 7 \\ \Rightarrow (\frac{n - 1}{2}) \cdot \frac{1}{{(2^{x})}^{3}} = 7 \end{array}$
Also, $^{n} C_{2} +^{n} C_{1} = 36$
$\begin{array}{l} \Rightarrow \frac{n (n - 1)}{2} + n = 36 \\ \Rightarrow n^{2} + n - 72 = 0 \\ \Rightarrow n = 8, - 9 \end{array}$
n = – 9 is not possible as in Eq. (1), n – 1 should be positive.
Substituting n = 8 in Eq. (1), we get
$\begin{matrix} 2^{3 x} = \frac{1}{2} = 2^{- 1} \\ \Rightarrow 3 x = - 1 \Rightarrow x = - \frac{1}{3} \end{matrix}$

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