First slide

Permutations

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 the third term is:

Moderate

A

455
B

105
C

605
D

120

Solution

$\begin{aligned} T_{r + 1} =^{n} C_{r} {(x^{3})}^{n - r} {(- \frac{1}{x^{2}})}^{r} \\ =^{n} C_{r} (- 1)^{r} x^{3 n - 5 r} \end{aligned}$
For coefficient of $x^{5},$ set $3 n - 5 r = 5$
$\Rightarrow r = (3 n - 10) / 5 = m$ (say)
Note that $l - m = 1$ .
We are given
$\begin{array}{l} ^{n} C_{l} (- 1)^{l} +^{n} C_{m} (- 1)^{m} = 0 \\ \Rightarrow^{n} C_{l} =^{n} C_{m} \Rightarrow l + m = n . \\ \Rightarrow \frac{3 n - 5}{5} + \frac{3 n - 10}{5} = n \\ \Rightarrow 6 n - 15 = 5 n \Rightarrow n = 15 . \end{array}$
coefficient of third term
$=^{15} C_{2} (- 1)^{2} = 105$ .

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