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JEE Main 202312 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
12 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2023 · 12 April · Morning Shift · Mathematics · Binomial Theorem

The sum, of the coefficients of the first 50 terms in the binomial...

The sum, of the coefficients of the first 5050 terms in the binomial expansion of ${\left(1 - x\right)}^{100}, is equal to

Options

  1. A101C50101{C}_{50}
  2. B99C4999{C}_{49}
  3. C−101C50{-}^{101}{C}_{50}
  4. D−99C49{-}^{99}{C}_{49}

Correct answer

D. −99C49{-}^{99}{C}_{49}

Step-by-step solution

We know that (1−x)100=C0100−C1100⋅x+C2100⋅x2−.......−C49100⋅x49+.......+C100100⋅x100{\left(1 - x\right)}^{100} = C0 100 - C1 100 \cdot x + C2 100 \cdot {x}^{2} - . . . . . . . - C49 100 \cdot {x}^{49} + . . . . . . . + C100 100 \cdot {x}^{100}

Now let, the sum of first fifty coefficients be SS.

⇒S=C0100−C1100+C2100−.......−C49100⇒ S = C0 100 - C1 100 + C2 100 - . . . . . . . - C49 100

Let us substitute x=1x = 1 in the expansion.

⇒(1−1)100=C0100−C1100⋅1+C2100⋅12−.......−C49100⋅149+.......+C100100⋅1100⇒ {\left(1 - 1\right)}^{100} = C0 100 - C1 100 \cdot 1 + C2 100 \cdot {1}^{2} - . . . . . . . - C49 100 \cdot {1}^{49} + . . . . . . . + C100 100 \cdot {1}^{100}

⇒0=C0100−C1100+C2100−.......−C49100+C50100−C51100+C52100.......+C100100⇒ 0 = C0 100 - C1 100 + C2 100 - . . . . . . . - C49 100 + C50 100 - C51 100 + C52 100 . . . . . . . + C100 100

There are total of 101101 terms and the middle term is T51=C50100{T}_{51} = C50 100.

As Crn=Cn−rnCr n = Cn - r n

⇒0=C0100−C1100+C2100−.......−C49100+C50100−C49100+C48100.......+C0100⇒ 0 = C0 100 - C1 100 + C2 100 - . . . . . . . - C49 100 + C50 100 - C49 100 + C48 100 . . . . . . . + C0 100

⇒0=C50100+C0100−C1100+C2100−.......−C49100−C49100+C48100.......+C0100⇒ 0 = C50 100 + C0 100 - C1 100 + C2 100 - . . . . . . . - C49 100 - C49 100 + C48 100 . . . . . . . + C0 100

⇒0=C50100+2(C0100−C1100+C2100−.......−C49100)⇒ 0 = C50 100 + 2\left(C0 100 - C1 100 + C2 100 - . . . . . . . - C49 100\right)

⇒−C501002=(C0100−C1100+C2100−.......−C49100)⇒ - \dfrac{C50 100}{2} = \left(C0 100 - C1 100 + C2 100 - . . . . . . . - C49 100\right)

⇒C0100−C1100+C2100−.......−C49100=−C501002⇒ C0 100 - C1 100 + C2 100 - . . . . . . . - C49 100 = - \dfrac{C50 100}{2}

Therefore, the required answer is −C501002=−12×10050C4999=−C4999- \dfrac{C50 100}{2} = - \dfrac{1}{2} \times \dfrac{100}{50} C49 99 = - C49 99.

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