If in the expansion of x3−1x2n the sum of the coefficients of x5 and x10 is 0, then the coefficient of x20 is:

A
$^{20} C_{6}$
B
$-^{20} C_{6}$
C
$^{15} C_{5}$
D
$-^{15} C_{5}$

Solution:

$(r + 1)$ th term in the expansion of

${(x^{3} - \frac{1}{x^{2}})}^{n}$ is

$T_{r + 1} =^{n} C_{r} {(x^{3})}^{n - r} {(- \frac{1}{x^{2}})}^{r}$

$=^{n} C_{r} x^{3 n - 5 r} (- 1)^{r}$

For coefficient of $x^{10}$ set $3 n - 5 r = 5$

$\Rightarrow r = \frac{3}{5} n - 1 = r_{1}$ (say)

Note that $r_{1} = r_{2} + 1$ .

We are given

$\begin{array}{l} ^{n} C_{r_{1}} (- 1)^{r_{1}} +^{n} C_{r_{2}} (- 1)^{r_{2}} = 0 \\ \Rightarrow^{n} C_{r_{2}} =^{n} C_{r_{2} + 1} [∵ r_{1} = r_{2} + 1] \\ \Rightarrow r_{2} + r_{2} + 1 = n \Rightarrow r_{2} = \frac{1}{2} (n - 1) \\ \Rightarrow \frac{3}{5} n - 2 = \frac{1}{2} n - \frac{1}{2} \Rightarrow \frac{1}{10} n = \frac{3}{2} \Rightarrow n = 15 \end{array}$

For coefficient of $x^{20}$ set $3 n - 5 r = 20$

$\Rightarrow 5 r = 45 - 20 = 25$ or $r = 5$ .

Thus, coefficient of $x^{20}$ is $-^{15} C_{5}$

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:

Solution:

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