First slide

Binomial theorem for positive integral Index

Question

If $a$ is real and the 4th term in the expansion of ${(a x + \frac{1}{x})}^{n}$ is $5 / 2$ , for each $x \in R - {0},$ then values of $n$ and $a$ are respectively

Moderate

A

$5, 1 / 2$
B

$6, - 1 / 2$
C

$3, 1 / 3$
D

$6, 1 / 2$

Solution

We have $\begin{aligned} T_{4} = T_{3} + 1 =^{n} C_{3} (a x)^{n - 3} {(\frac{1}{x})}^{3} \\ =^{n} C_{3} a^{n - 3} x^{n - 6} = 5 / 2 \end{aligned}$
As this is true for each $x \in R - {0},$ we get
$n - 6 = 0$ and $^{n} C_{3} a^{n - 3} = 5 / 2$
$\Rightarrow$ $n = 6$ and $^{6} C_{3} a^{3} = 5 / 2$
$\begin{array}{l} ∴ a^{3} = \frac{5}{2} \times \frac{3! 3!}{6!} = \frac{5}{2} \times \frac{1}{20} = \frac{1}{8} \\ \Rightarrow a = 1 / 2 . \end{array}$
Thus, $n = 6, a = 1 / 2 .$

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