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Binomial Theorem question

2019 · 12 Apr · Shift 1 · Q42
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  5. /2019 · 12 Apr · Shift 1 · Q42

Binomial Theorem question

2019 · 12 Apr · Shift 1 · Q42

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
The coefficient of x18 in the product (1 + x) (1 – x)10 (1 + x + x2)9 is :
  1. A
    126
  2. B
    - 84
  3. C
    - 126
  4. D
    84
View written solutionFree

Correct answer: D

  1. Simplify the expression

We need the coefficient of x18x^{18}x18 in

(1+x)(1−x)10(1+x+x2)9.(1+x)(1-x)^{10}(1+x+x^2)^9.(1+x)(1−x)10(1+x+x2)9.

Use the identity

1+x+x2=1−x31−x.1+x+x^2 = \frac{1-x^3}{1-x}.1+x+x2=1−x1−x3​.

So,

(1+x)(1−x)10(1+x+x2)9;=(1+x)(1−x)10(1−x31−x)9.(1+x)(1-x)^{10}(1+x+x^2)^9 ;= (1+x)(1-x)^{10}\left(\frac{1-x^3}{1-x}\right)^9.(1+x)(1−x)10(1+x+x2)9;=(1+x)(1−x)10(1−x1−x3​)9.

Thus,

=(1+x)(1−x)(1−x3)9=(1−x2)(1−x3)9.= (1+x)(1-x)(1-x^3)^9 = (1-x^2)(1-x^3)^9.=(1+x)(1−x)(1−x3)9=(1−x2)(1−x3)9.

So the problem reduces to finding the coefficient of x18x^{18}x18 in

(1−x2)(1−x3)9.(1-x^2)(1-x^3)^9.(1−x2)(1−x3)9.
  1. Expand (1−x3)9(1-x^3)^9(1−x3)9

By binomial theorem,

(1−x3)9=∑k=09(9k)(−1)kx3k.(1-x^3)^9 = \sum_{k=0}^9 \binom{9}{k}(-1)^k x^{3k}.(1−x3)9=k=0∑9​(k9​)(−1)kx3k.

Now multiply by (1−x2)(1-x^2)(1−x2):

(1−x2)(1−x3)9=(1−x3)9−x2(1−x3)9.(1-x^2)(1-x^3)^9 = (1-x^3)^9 - x^2(1-x^3)^9.(1−x2)(1−x3)9=(1−x3)9−x2(1−x3)9.

We want the coefficient of x18x^{18}x18.

  1. Find contribution to x18x^{18}x18

From (1−x3)9(1-x^3)^9(1−x3)9

We need

3k=18  ⟹  k=6.3k=18 \implies k=6.3k=18⟹k=6.

So the coefficient is

(96)(−1)6=(96)=84.\binom{9}{6}(-1)^6 = \binom{9}{6} = 84.(69​)(−1)6=(69​)=84.

From −x2(1−x3)9-x^2(1-x^3)^9−x2(1−x3)9

We would need a term x16x^{16}x16 in (1−x3)9(1-x^3)^9(1−x3)9. But powers in (1−x3)9(1-x^3)^9(1−x3)9 are only multiples of 333, and 161616 is not a multiple of 333. So this contributes

0.0.0.

Hence the required coefficient is

84.84.84.
  1. Check options
  • A: 126126126 ❌
  • B: −84-84−84 ❌
  • C: −126-126−126 ❌
  • D: 848484 ✅

So the correct option is D.

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