In the expansion of x3−1x2n,n∈N, if the sum of the coefficients of x5 and x10is 0, then n is

# In the expansion of ${\left({\mathrm{x}}^{3}-\frac{1}{{\mathrm{x}}^{2}}\right)}^{\mathrm{n}},\mathrm{n}\in \mathrm{N}$, if the sum of the coefficients of is 0, then n is

1. A

25

2. B

20

3. C

15

4. D

None of these

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### Solution:

${\left({\mathrm{x}}^{3}-\frac{1}{{\mathrm{r}}^{2}}\right)}^{\mathrm{n}}$
General term ${\mathrm{T}}_{\mathrm{r}+1}=\frac{\mathrm{n}!}{\mathrm{r}!\left(\mathrm{n}-\mathrm{r}\right)!}\left(-1{\right)}^{\mathrm{n}-\mathrm{r}}{\mathrm{x}}^{5\mathrm{n}-2\mathrm{n}}$
If
If

Given that sum of x5 and x10 is zero.
$⇒\frac{5\mathrm{k}!}{\left(2\mathrm{k}+1\right)!\left(3\mathrm{k}-1\right)!}-\frac{5\mathrm{k}!}{\left(2\mathrm{k}+2\right)!\left(3\mathrm{k}-2\right)!}=0$
or  $\frac{1}{3\mathrm{k}-1}-\frac{1}{2\mathrm{k}+2}=0$
or   $\mathrm{k}=3⇒\mathrm{n}=15$

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