First slide

Binomial theorem for positive integral Index

Question

In the expansion of ${(\frac{1}{x^{2}} - x^{3})}^{n}, n \in N$ , if the sum of the coefficients of x⁵ and x¹⁰ is 0, then n is

Moderate

A

25
B

20
C

15
D

5

Solution

${(\frac{1}{x^{2}} - x^{3})}^{n}$
General term $T_{r + 1} = n_{c_{r}} {(\frac{1}{x^{2}})}^{n - r} {(- x^{3})}^{r} = \frac{n!}{r! (n - r)!} (- 1)^{n - r} x^{5 r - 2 n}$
for coefficient $x^{5}$ 5r - 2n = 5, then 5r = 2n + 5 or $r = \frac{2 n}{5} + 1$ …………(I)
for coefficient $x^{10}$ 5r - 2n = 10, then 5r = 2n + 10 or $r = \frac{2 n}{5} + 2$ ……………………(II)
$x^{5}$ and x¹⁰ terms occurs if n = 5k
in (I) put n = 5k $\Rightarrow$ r = 2k+1
in (II) put n = 5k $\Rightarrow$ r = 2k+2
Given that sum of x⁵ and x¹⁰ is zero.
$\begin{array}{l} \Rightarrow \frac{5 k!}{(2 k + 1)! (3 k - 1)!} - \frac{5 k!}{(2 k + 2)! (3 k - 2)!} = 0 \\ or \frac{1}{3 k - 1} - \frac{1}{2 k + 2} = 0 \\ or k = 3 \Rightarrow n = 15 \end{array}$

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