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Binomial Theorem question

2004 · Shift 0 · Q106
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  5. /2004 · Shift 0 · Q106

Binomial Theorem question

2004 · Shift 0 · Q106

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
The coefficient of xn{x^n}xn in expansion of (1+x)(1−x)n\left( {1 + x} \right){\left( {1 - x} \right)^n}(1+x)(1−x)n is
  1. A
    (−1)n−1n{\left( { - 1} \right)^{n - 1}}n(−1)n−1n
  2. B
    (−1)n(1−n){\left( { - 1} \right)^n}\left( {1 - n} \right)(−1)n(1−n)
  3. C
    (−1)n−1(n−1)2{\left( { - 1} \right)^{n - 1}}{\left( {n - 1} \right)^2}(−1)n−1(n−1)2
  4. D
    (n−1)\left( {n - 1} \right)(n−1)
View written solutionFree

Correct answer: B

  1. We need the coefficient of xnx^nxn in
(1+x)(1−x)n.(1+x)(1-x)^n.(1+x)(1−x)n.
  1. Expand strategically:
(1+x)(1−x)n=(1−x)n+x(1−x)n.(1+x)(1-x)^n=(1-x)^n+x(1-x)^n.(1+x)(1−x)n=(1−x)n+x(1−x)n.

So the coefficient of xnx^nxn will be:

  • coefficient of xnx^nxn in (1−x)n(1-x)^n(1−x)n
  • plus coefficient of xnx^nxn in x(1−x)nx(1-x)^nx(1−x)n
  1. Coefficient of xnx^nxn in (1−x)n(1-x)^n(1−x)n: Using binomial expansion,
(1−x)n=∑k=0n(nk)(−x)k.(1-x)^n=\sum_{k=0}^n \binom{n}{k}(-x)^k.(1−x)n=k=0∑n​(kn​)(−x)k.

Hence coefficient of xnx^nxn is

(−1)n(nn)=(−1)n.(-1)^n\binom{n}{n}=(-1)^n.(−1)n(nn​)=(−1)n.
  1. Coefficient of xnx^nxn in x(1−x)nx(1-x)^nx(1−x)n: This equals coefficient of xn−1x^{n-1}xn−1 in (1−x)n(1-x)^n(1−x)n. Now coefficient of xn−1x^{n-1}xn−1 in (1−x)n(1-x)^n(1−x)n is
(−1)n−1(nn−1)=(−1)n−1n.(-1)^{n-1}\binom{n}{n-1}=(-1)^{n-1}n.(−1)n−1(n−1n​)=(−1)n−1n.

So coefficient of xnx^nxn in x(1−x)nx(1-x)^nx(1−x)n is

(−1)n−1n.(-1)^{n-1}n.(−1)n−1n.
  1. Add both contributions:
(−1)n+(−1)n−1n.(-1)^n+(-1)^{n-1}n.(−1)n+(−1)n−1n.

Factor out (−1)n(-1)^n(−1)n:

(−1)n+(−1)n−1n=(−1)n−(−1)nn=(−1)n(1−n).(-1)^n+(-1)^{n-1}n = (-1)^n-(-1)^n n = (-1)^n(1-n).(−1)n+(−1)n−1n=(−1)n−(−1)nn=(−1)n(1−n).
  1. Therefore, the required coefficient is
(−1)n(1−n).\boxed{(-1)^n(1-n)}.(−1)n(1−n)​.
  1. Compare with options:
  • A: (−1)n−1n(-1)^{n-1}n(−1)n−1n ❌
  • B: (−1)n(1−n)(-1)^n(1-n)(−1)n(1−n) ✅
  • C: (−1)n−1(n−1)2(-1)^{n-1}(n-1)^2(−1)n−1(n−1)2 ❌
  • D: (n−1)(n-1)(n−1) ❌

So the correct option is B.

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