First slide

Binomial theorem for positive integral Index

Question

Suppose $a \in R .$ If the coefficient of $x^{5}$ in the expansion of ${(a x + \frac{1}{x^{3}})}^{17}$ is $680,$ then $a$ is equal to

Moderate

A

$\pm 2$
B

$\pm 1$
C

$\pm 1 / 2$
D

$\pm 1 / 3$

Solution

$\begin{aligned} T_{r + 1} =^{17} C_{r} (a x)^{17 - r} {(\frac{1}{x^{3}})}^{r} \\ =^{17} C_{r} a^{17 - r} x^{17 - 4 r} \end{aligned}$
For the coefficient of $x^{5},$ we set
$\begin{array}{l} 17 - 4 r = 5 or r = 3 . \\ ∴^{17} C_{3} a^{14} = 680 \\ \Rightarrow a^{14} = \frac{680 \times 3! 14!}{17!} = 1 \\ \Rightarrow a = \pm 1 . \end{array}$

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