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Binomial Theorem question

2006 · Shift 0 · Q73
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  5. /2006 · Shift 0 · Q73

Binomial Theorem question

2006 · Shift 0 · Q73

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If the expansion in powers of xxx of the function 1(1−ax)(1−bx){1 \over {\left( {1 - ax} \right)\left( {1 - bx} \right)}}(1−ax)(1−bx)1​ is a0+a1x+a2x2+a3x3.....{a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}.....a0​+a1​x+a2​x2+a3​x3..... then an{a_n}an​ is
  1. A
    bn−anb−a{{{b^n} - {a^n}} \over {b - a}}b−abn−an​
  2. B
    an−bnb−a{{{a^n} - {b^n}} \over {b - a}}b−aan−bn​
  3. C
    an+1−bn+1b−a{{{a^{n + 1}} - {b^{n + 1}}} \over {b - a}}b−aan+1−bn+1​
  4. D
    bn+1−an+1b−a{{{b^{n + 1}} - {a^{n + 1}}} \over {b - a}}b−abn+1−an+1​
View written solutionFree

Correct answer: D

  1. We need the coefficient of xnx^nxn in
1(1−ax)(1−bx).\frac{1}{(1-ax)(1-bx)}.(1−ax)(1−bx)1​.
  1. Use partial fractions:
1(1−ax)(1−bx)=A1−ax+B1−bx.\frac{1}{(1-ax)(1-bx)}=\frac{A}{1-ax}+\frac{B}{1-bx}.(1−ax)(1−bx)1​=1−axA​+1−bxB​.

Multiplying by (1−ax)(1−bx)(1-ax)(1-bx)(1−ax)(1−bx),

1=A(1−bx)+B(1−ax).1=A(1-bx)+B(1-ax).1=A(1−bx)+B(1−ax).

Comparing coefficients,

A+B=1,Ab+Ba=0.A+B=1, \qquad Ab+Ba=0.A+B=1,Ab+Ba=0.

A quicker standard decomposition is

1(1−ax)(1−bx)=1b−a(b1−bx−a1−ax).\frac{1}{(1-ax)(1-bx)}=\frac{1}{b-a}\left(\frac{b}{1-bx}-\frac{a}{1-ax}\right).(1−ax)(1−bx)1​=b−a1​(1−bxb​−1−axa​).

Let us verify:

1b−a(b(1−ax)−a(1−bx)(1−ax)(1−bx))=1b−a⋅b−a(1−ax)(1−bx)=1(1−ax)(1−bx).\frac{1}{b-a}\left(\frac{b(1-ax)-a(1-bx)}{(1-ax)(1-bx)}\right) =\frac{1}{b-a}\cdot\frac{b-a}{(1-ax)(1-bx)} =\frac{1}{(1-ax)(1-bx)}.b−a1​((1−ax)(1−bx)b(1−ax)−a(1−bx)​)=b−a1​⋅(1−ax)(1−bx)b−a​=(1−ax)(1−bx)1​.
  1. Now expand each term as a geometric series:
11−bx=1+bx+b2x2+b3x3+⋯\frac{1}{1-bx}=1+bx+b^2x^2+b^3x^3+\cdots1−bx1​=1+bx+b2x2+b3x3+⋯

and

11−ax=1+ax+a2x2+a3x3+⋯\frac{1}{1-ax}=1+ax+a^2x^2+a^3x^3+\cdots1−ax1​=1+ax+a2x2+a3x3+⋯

So,

b1−bx=b+b2x+b3x2+b4x3+⋯\frac{b}{1-bx}=b+b^2x+b^3x^2+b^4x^3+\cdots1−bxb​=b+b2x+b3x2+b4x3+⋯

and

a1−ax=a+a2x+a3x2+a4x3+⋯\frac{a}{1-ax}=a+a^2x+a^3x^2+a^4x^3+\cdots1−axa​=a+a2x+a3x2+a4x3+⋯
  1. Therefore,
1(1−ax)(1−bx)=1b−a[(b−a)+(b2−a2)x+(b3−a3)x2+⋯].\frac{1}{(1-ax)(1-bx)} =\frac{1}{b-a}\Big[(b-a)+(b^2-a^2)x+(b^3-a^3)x^2+\cdots\Big].(1−ax)(1−bx)1​=b−a1​[(b−a)+(b2−a2)x+(b3−a3)x2+⋯].

Hence the coefficient of xnx^nxn is

an=bn+1−an+1b−a.a_n=\frac{b^{n+1}-a^{n+1}}{b-a}.an​=b−abn+1−an+1​.
  1. Match with options:
  • A: bn−anb−a\dfrac{b^n-a^n}{b-a}b−abn−an​
  • B: an−bnb−a\dfrac{a^n-b^n}{b-a}b−aan−bn​
  • C: an+1−bn+1b−a\dfrac{a^{n+1}-b^{n+1}}{b-a}b−aan+1−bn+1​
  • D: bn+1−an+1b−a\dfrac{b^{n+1}-a^{n+1}}{b-a}b−abn+1−an+1​

So the correct option is D.

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