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Binomial Theorem question

2020 · 9 Jan · Shift 1 · Q32
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  5. /2020 · 9 Jan · Shift 1 · Q32

Binomial Theorem question

2020 · 9 Jan · Shift 1 · Q32

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
The coefficient of x4 is the expansion of (1 + x + x2)10 is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 615

  1. We need the coefficient of x4x^4x4 in
(1+x+x2)10.(1+x+x^2)^{10}.(1+x+x2)10.
  1. Rewrite the expression:
1+x+x2=1−x31−x.1+x+x^2=\frac{1-x^3}{1-x}.1+x+x2=1−x1−x3​.

Hence,

(1+x+x2)10=(1−x31−x)10=(1−x3)10(1−x)−10.(1+x+x^2)^{10}=\left(\frac{1-x^3}{1-x}\right)^{10}=(1-x^3)^{10}(1-x)^{-10}.(1+x+x2)10=(1−x1−x3​)10=(1−x3)10(1−x)−10.
  1. Now expand each factor only as much as needed for the coefficient of x4x^4x4.

For (1−x3)10(1-x^3)^{10}(1−x3)10:

(1−x3)10=1−10x3+⋯(1-x^3)^{10}=1-10x^3+\cdots(1−x3)10=1−10x3+⋯

Terms involving x6,x9,…x^6, x^9, \dotsx6,x9,… cannot contribute to x4x^4x4 when multiplied by nonnegative powers from the second factor.

For (1−x)−10(1-x)^{-10}(1−x)−10, using

(1−x)−10=∑n=0∞(n+99)xn,(1-x)^{-10}=\sum_{n=0}^{\infty} \binom{n+9}{9}x^n,(1−x)−10=n=0∑∞​(9n+9​)xn,

we get:

  • coefficient of x4x^4x4 is (139)=(134)=715\binom{13}{9}=\binom{13}{4}=715(913​)=(413​)=715,
  • coefficient of x1x^1x1 is (109)=10\binom{10}{9}=10(910​)=10.
  1. Therefore, coefficient of x4x^4x4 in the product is
1⋅[x4](1−x)−10+(−10x3)⋅[x1](1−x)−10.1\cdot [x^4](1-x)^{-10}+(-10x^3)\cdot [x^1](1-x)^{-10}.1⋅[x4](1−x)−10+(−10x3)⋅[x1](1−x)−10.

So,

[x4](1+x+x2)10=715−10⋅10=715−100=615.[x^4](1+x+x^2)^{10}=715-10\cdot 10=715-100=615.[x4](1+x+x2)10=715−10⋅10=715−100=615.
  1. Therefore, the required coefficient is
615.\boxed{615}.615​.
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