First slide

Binomial theorem for positive integral Index

Question

If in the expansion of ${(x^{3} - \frac{1}{x^{2}})}^{n}$ the
sum of the coefficients of $x^{5}$ and $x^{10}$ is $0$
then the coefficient of $x^{20}$ is

Moderate

A

$_{}^{20} C_{6}$
B

$-^{20} C_{6}$
C

$^{15} C_{5}$
D

$^{- 15} C_{5}$

Solution

$(r + 1) t h$ term in the expansion of
$\begin{matrix} {(x^{3} - \frac{1}{x^{2}})}^{n} is \\ T_{r + 1} =^{n} C_{r} {(x^{3})}^{n - r} {(- \frac{1}{x^{2}})}^{r} \end{matrix}$
$=^{n} C_{r} x^{3 n - 5 r} (- 1)^{r}$
For coefficient of $r^{10},$ set $3 n - 5 r = 10$
$\Rightarrow r = \frac{3}{5} n - 1 = r_{1}$ (say)
Note that, $r_{1} = r_{2} + 1$
We are given
$\begin{array}{l} ^{n} C_{r_{1}} (- 1)^{r_{1}} +^{n} C_{r_{2}} (- 1)^{r_{2}} = 0 \\ ^{n} C_{r_{2}} =^{n} C_{r_{2} + 1} [∵ r_{1} = r_{2} + 1] \\ \Rightarrow r_{2} + r_{2} + 1 = n \Rightarrow r_{2} = \frac{1}{2} (n - 1) \\ \Rightarrow \frac{3}{5} n - 2 = \frac{1}{2} n - \frac{1}{2} \Rightarrow \frac{1}{10} n = \frac{3}{2} \Rightarrow n = 15 \end{array}$
For coefficient of $x^{20}$ set $3 n - 5 r = 20$
$5 r = 45 - 20 = 25$ or $r = 5$
Thus, coefficient of $x^{20}$ is $-^{15} C_{5}$

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