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JEE Main 202325 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
25 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2023 · 25 January · Evening Shift · Mathematics · Binomial Theorem

∑ _ k = 0 ^ 6 C3 51 - k is equal to

∑k=06C351−k\mathop{∑}_{k = 0}^{6} C3 51 - k is equal to

Options

  1. AC451−C445C4 51 - C4 45
  2. BC351−C345C3 51 - C3 45
  3. CC452−C445C4 52 - C4 45
  4. DC352−C345C3 52 - C3 45

Correct answer

C. C452−C445C4 52 - C4 45

Step-by-step solution

Given,

∑r=06C351−k\mathop{∑}_{r = 0}^{6} C3 51 - k

=C351+C350+C349+C348+C347+C346+C345= C3 51 + C3 50 + C3 49 + C3 48 + C3 47 + C3 46 + C3 45

Now we know that Crn+Cr+1n=Cr+1n+1Cr n + Cr + 1 n = Cr + 1 n + 1 or Crn=Cr+1n+1−Cr+1nCr n = Cr + 1 n + 1 - Cr + 1 n

Now using the above formula in given expression we get,

∑r=06C351−k=C351+C350+C349+C348+C347+C346+C446⏟−C445\mathop{∑}_{r = 0}^{6} C3 51 - k = C3 51 + C3 50 + C3 49 + C3 48 + C3 47 + \mathop{\mathop{C3 46 + C4 46}_{⏟}}_{} - C4 45

⇒∑r=06C351−k=C351+C350+C349+C348+C347+C447⏟−C445⇒ \mathop{∑}_{r = 0}^{6} C3 51 - k = C3 51 + C3 50 + C3 49 + C3 48 + \mathop{\mathop{C3 47 + C4 47}_{⏟}}_{} - C4 45

⇒∑r=06C351−k=C351+C350+C349+C348+C448−C445⇒ \mathop{∑}_{r = 0}^{6} C3 51 - k = C3 51 + C3 50 + C3 49 + C3 48 + C4 48 - C4 45

⇒∑r=06C351−k=C351+C350+C349+C449−C445⇒ \mathop{∑}_{r = 0}^{6} C3 51 - k = C3 51 + C3 50 + C3 49 + C4 49 - C4 45

⇒∑r=06C351−k=C351+C350+C450−C445⇒ \mathop{∑}_{r = 0}^{6} C3 51 - k = C3 51 + C3 50 + C4 50 - C4 45

⇒∑r=06C351−k=C351+C451−C445⇒ \mathop{∑}_{r = 0}^{6} C3 51 - k = C3 51 + C4 51 - C4 45

⇒∑r=06C351−k=C452−C445⇒ \mathop{∑}_{r = 0}^{6} C3 51 - k = C4 52 - C4 45.

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