First slide

Binomial theorem for positive integral Index

Question

The coefficient of x¹⁸ in the product $(1 + x) (1 - x)^{10} {(1 + x + x^{2})}^{9} is$

Moderate

A

84
B

-126
C

-84
D

126

Solution

Given expression is $(1 + x) (1 - x)^{10} {(1 + x + x^{2})}^{9}$
$= (1 + x) (1 - x) {[(1 - x) (1 + x + x^{2})]}^{9} = (1 - x^{2}) {(1 - x^{3})}^{9}$
Now, coefficient of x¹⁸ in the product
$(1 + x) (1 - x)^{10} {(1 + x + x^{2})}^{9}$
=coefficient of y¹⁸ in the product $(1 - x^{2}) {(1 - x^{3})}^{9}$
=coefficient of y¹⁸ in (1 - x³ )⁹- coefficient of x¹⁶
in (1 - x³)⁹
Since, (r + 1)th term in the expansion of
${(1 - x^{3})}^{9} {is}^{9} C_{r} {(- x^{3})}^{r} =^{9} C_{r} (- 1)^{r} x^{3 r}$
Now, for x¹⁸, 3r = 18 $\Rightarrow$ r = 6
and for x¹⁸ , $3 r = 18 \Rightarrow r = 6$
and for x¹⁶, $3 r = 16 \Rightarrow r = \frac{16}{3} \notin N$
.'. Required coefficient is $^{9} C_{6} = \frac{9!}{6! 3!} = \frac{9 \times 8 \times 7}{3 \times 2} = 84$

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