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Binomial Theorem

JEE Main 2024 (Online) 30th January Evening ShiftNumericalHard📈 Topic in 43% papers+4
Ques. 1JEE Main 2024, 30 January (Evening Shift)

Let = k=0n ( ( n Ck )2k+1 ) and = k=0n-1 ( n Ck n Ck+1k+2 ). If 5 =6 , then n equals .

Answer

10

Solution

∵ nCrr+1= n+1Cr+1n+1

= 1n+1 k=0nnCn-k n+1Ck+1= 2n+1Cn+1n+1

= 1n+1 k=0n-1nCn-k n+1Ck+2= 2n+1Cn+2n+1

∵ mCr nCs-r=m+nCs (lower indices add to n+1 and n+2)

= 2n+1Cn+22n+1Cn+1= nn+2

5 =6 nn+2= 56 n=10

Answer: 10

Keep practising — Binomial Theorem

Ques. 2JEE Main 2024, 29 January (Morning Shift)

If 11C12+ 11C23+ + 11C910= nm with (n,m)=1, then n+m is equal to .

Ques. 3JEE Main 2024, 29 January (Evening Shift)

Remainder when 643232 is divided by 9 is equal to .

Ques. 4JEE Main 2024, 27 January (Morning Shift)

If A denotes the sum of all the coefficients in the expansion of (1-3x+10x2 )n and B denotes the sum of all the coefficients in the expansion of (1+x2 )n, then :

  1. B=A3
  2. 3A=B
  3. A=3B
  4. A=B3
Ques. 5JEE Main 2024, 27 January (Evening Shift)

The coefficient of x2012 in the expansion of (1-x)2008 (1+x+x2 )2007 is equal to .

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