First slide

Binomial theorem for positive integral Index

Question

The term independent of $x$ in the expansion of ${(a x + \frac{b}{x})}^{14}$ is

Easy

A

$14! a^{7} b^{7}$
B

$\frac{14!}{7!} a^{7} b^{7}$
C

$\frac{14!}{7! 7!} a^{7} b^{7}$
D

$\frac{14!}{7! 7! 7!} a^{7} b^{7}$

Solution

Given expansion is ${(a x + \frac{b}{x})}^{14}$
We have general term in the expansion ${(x + a)}^{n}$
$(∴ T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$ be the expansion of ${(x + a)}^{n})$
$T_{r + 1} =^{14} C_{r} {(a x)}^{14 - r} {(\frac{b}{x})}^{r}$
$T_{r + 1} =^{14} C_{r} {(a)}^{14 - r} {(x)}^{14 - r} {(b)}^{r} {(x^{- 1})}^{r}$ $(∴ \frac{x^{a}}{x^{b}} = x^{a - b})$
   $T_{r + 1} =^{14} C_{r} a^{14 - r} b^{r} x^{14 - 2 r} ............ (1)$
$x^{14 - 2 r}$ compare with $x^{0}$ because of independent term
$\Rightarrow x^{14 - 2 r} = x^{0}$
This will be independent of $x$ if $14 - 2 r = 0$
$∵ r = 7$ this value substitute in Eqn (1)
$∴$ The term without $x$ is equal to
$T_{8} =^{14} C_{7} a^{7} b^{7} = \frac{14!}{7! 7!} a^{7} b^{7}$

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