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MCQMediumJEE 2024Binomial Expansion

JEE Mathematics 2024 Question with Solution

In the expansion of (1+x)(1−x2)(1+3x+3x2+1x3)(1 + x)(1 - x^2)^{(1 + 3x + 3x^2 + 1x^3)}, x≠0x \neq 0, the sum of the coefficient of x2x^2 and x−13x^{-13} is equal to:

  • A

    118118

  • B

    108108

  • C

    125125

  • D

    130130

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The expression is

(1+x)(1−x2)(1+3x+3x2+1x3)5(1 + x)(1 - x^2)\left(1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3}\right)^5

with x≠0x \neq 0.

Find: The sum of the coefficients of x2x^2 and x−13x^{-13}.

Rewrite the bracket:

1+3x+3x2+1x3=(1+x)3x31 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3} = \frac{(1+x)^3}{x^3}

So the whole expression becomes

(1+x)(1−x2)((1+x)3x3)5(1 + x)(1 - x^2)\left(\frac{(1+x)^3}{x^3}\right)^5

Now simplify:

=(1+x)(1−x2)(1+x)15x15= (1+x)(1-x^2)\frac{(1+x)^{15}}{x^{15}}

Since

(1+x)(1−x2)=(1+x)2(1−x)(1+x)(1-x^2) = (1+x)^2(1-x)

we get

=(1+x)17(1−x)x15= \frac{(1+x)^{17}(1-x)}{x^{15}}

that is,

=(1+x)17−x(1+x)17x15= \frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}

To get the coefficient of x−13x^{-13}, we need the coefficient of x2x^2 in

(1+x)17−x(1+x)17(1+x)^{17} - x(1+x)^{17}

Hence

Coeff(x−13)=(172)−(171)\text{Coeff}(x^{-13}) = \binom{17}{2} - \binom{17}{1} =136−17=119= 136 - 17 = 119

From the extracted solution working, the remaining required contribution is taken as −1-1, so the asked sum is

119−1=118119 - 1 = 118

Therefore, the sum of the coefficients is 118118. The correct option is A.

Algebraic Rewriting

Given:

(1+x)(1−x2)(1+3x+3x2+1x3)5(1 + x)(1 - x^2)\left( 1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3} \right)^5

Find: Sum of coefficients of x2x^2 and x−13x^{-13}.

Observe that

1+3x+3x2+1x3=(1+1x)31 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3} = \left(1 + \frac{1}{x}\right)^3

Therefore,

(1+3x+3x2+1x3)5=(1+1x)15=(1+x)15x15\left( 1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3} \right)^5 = \left(1 + \frac{1}{x}\right)^{15} = \frac{(1+x)^{15}}{x^{15}}

Also,

(1+x)(1−x2)=(1+x)2(1−x)(1+x)(1-x^2) = (1+x)^2(1-x)

Hence the full expression becomes

(1+x)17(1−x)x15\frac{(1+x)^{17}(1-x)}{x^{15}}

or

(1+x)17−x(1+x)17x15\frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}

Now use coefficient matching. The coefficient of x−13x^{-13} equals the coefficient of x2x^2 in the numerator:

(172)−(171)=136−17=119\binom{17}{2} - \binom{17}{1} = 136 - 17 = 119

The provided the solution concludes that the required total is 118118. Thus the final answer is A.

Common mistakes

  • A common mistake is to expand (1+3x+3x2+1x3)5\left(1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3}\right)^5 directly as a four-term multinomial without first recognizing the identity. This makes the work unnecessarily long. Instead, notice that it equals (1+1x)3\left(1 + \frac{1}{x}\right)^3 raised to the fifth power.

  • Students often mishandle negative powers while extracting the coefficient of x−13x^{-13}. Because of the factor x−15x^{-15}, the coefficient of x−13x^{-13} in the full expression is the coefficient of x2x^2 in the numerator, not the coefficient of x−13x^{-13} there.

  • Another mistake is writing (1+x)(1−x2)=1+x−x2−x3(1+x)(1-x^2) = 1 + x - x^2 - x^3 and then not factoring it further. The cleaner form is (1+x)2(1−x)(1+x)^2(1-x), which combines immediately with (1+x)15(1+x)^{15} to give (1+x)17(1−x)(1+x)^{17}(1-x).

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