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Binomial Theorem

JEE Main 2024 (Online) 30th January Morning ShiftNumericalEasy📈 Topic in 21% papers+4
Ques. 1JEE Main 2024, 30 January (Morning Shift)

Number of integral terms in the expansion of \7 ( 12 )+11 ( 16 ) \824 is equal to .

Answer

138

Solution

Tr+1=824Cr 7 824-r2 11 r6

Integral r6 and 824-r2 are integers r is a multiple of 6 (then r is even too)

r=0, 6, 12, , 822 (0 r 824)

Number of terms = 8226+1=138

Answer: 138

Keep practising — Binomial Theorem

Ques. 2JEE Main 2024, 30 January (Evening Shift)

Suppose 2-p, p, 2- , are the coefficients of four consecutive terms in the expansion of (1+x)n. Then the value of p2- 2+6 +2p equals

  1. 8
  2. 4
  3. 6
  4. 10
Ques. 3JEE 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 .

Ques. 4JEE Main 2024, 29 January (Morning Shift)

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

Ques. 5JEE Main 2024, 29 January (Evening Shift)

Remainder when 643232 is divided by 9 is equal to .

Ques. 6JEE 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. 7JEE 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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