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JEE Main 202324 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2023 · 24 January · Morning Shift · Mathematics · Binomial Theorem

The value of ∑ _ r = 0 ^ 22 Cr 22 · Cr 23 is

The value of ∑r=022Cr22⋅Cr23\mathop{∑}_{r = 0}^{22} Cr 22 \cdot Cr 23 is

Options

  1. AC2345C23 45
  2. BC2344C23 44
  3. CC2445C24 45
  4. DC2244C22 44

Correct answer

A. C2345C23 45

Step-by-step solution

Let, S=∑r=022Cr22⋅Cr23S = \mathop{∑}_{r = 0}^{22} Cr 22 \cdot Cr 23

Now using Crn=Cn−rnCr n = Cn - r n we get,

S=∑r=022Cr22⋅C23−r23S = \mathop{∑}_{r = 0}^{22} Cr 22 \cdot C23 - r 23

⇒S=C022⋅C2323+C122⋅C2223+............C2222⋅C123⇒ S = C0 22 \cdot C23 23 + C1 22 \cdot C22 23 + . . . . . . . . . . . . C22 22 \cdot C1 23

Which is sum of coefficient of x23intheexpansionof{x}^{23} in the expansion of{\left(1 + x\right)}^{22} \cdot {\left(1 + x\right)}^{23} or (1+x)23+22=(1+x)45{\left(1 + x\right)}^{23 + 22} = {\left(1 + x\right)}^{45}

Hence, ∑r=022Cr22⋅C23−r23=C2345\mathop{∑}_{r = 0}^{22} Cr 22 \cdot C23 - r 23 = C23 45

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