First slide

Binomial theorem for positive integral Index

Question

The sum of the coefficients of $x^{32}$ and $x^{- 17}$ in ${(x^{4} - \frac{1}{x^{3}})}^{15}$ is

Moderate

A

1
B

–1
C

2
D

0

Solution

Given expansion ${(x^{4} - \frac{1}{x^{3}})}^{15}$ is
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 ${(x + a)}^{n})$
$T_{r + 1} =^{15} C_{r} {(x^{4})}^{15 - r} {(\frac{- 1}{x^{3}})}^{r}$
$T_{r + 1} =^{15} C_{r} {(x)}^{60 - 7 r} {(- 1)}^{r}$
${(x)}^{60 - 7 r}$ compare with $x^{32}$
$\Rightarrow x^{60 - 7 r} = x^{32}$
$60 - 7 r = 32 \Rightarrow r = 4$
$\Rightarrow x^{60 - 7 r} = x^{- 17}$
$60 - 7 r = - 17 \Rightarrow r = 11$
The sum of the coefficients of $x^{32}$ and $x^{- 17}$ is
$=^{15} C_{4} +^{15} C_{11} {(- 1)}^{11} (∴^{15} C_{4} =^{15} C_{11})$
$\Rightarrow^{15} C_{4} -^{15} C_{11} = 0$

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