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Binomial Theorem

JEE Main 2024 (Online) 9th April Morning ShiftNumericalEasy📈 Topic in 10% papers+4
Ques. 1JEE Main 2024, 9 April (Morning Shift)

The remainder when 4282024 is divided by 21 is .

Answer

1

Solution

428 = 21 20 + 8, so 4282024 = (21 20 + 8)2024 leaves the same remainder as 82024 (all other binomial terms contain 21).

Now 82 = 64 = 63 + 1 = 21 3 + 1

82024 = (82)1012 = (1 + 63)1012 = 1 + 63 ( )

Hence the remainder is 1.

Answer: 1

Keep practising — Binomial Theorem

Ques. 2JEE Main 2024, 9 April (Evening Shift)

The sum of the coefficients of x2/3 and x-2/5 in the binomial expansion of (x2/3 + 12x-2/5 )9 is

  1. 19/4
  2. 69/16
  3. 63/16
  4. 21/4
Ques. 3JEE Main 2024, 8 April (Morning Shift)

Let = r=0n (4r2 + 2r + 1 )nCr and = ( r=0n nCrr + 1 ) + 1n + 1. If 140 < 2 < 281, then the value of n is .

Ques. 4JEE Main 2024, 8 April (Evening Shift)

If the term independent of x in the expansion of ( ax2 + 12x3 )10 is 105, then a2 is equal to :

  1. 6
  2. 4
  3. 2
  4. 9
Ques. 5JEE Main 2024, 6 April (Morning Shift)

If the second, third and fourth terms in the expansion of (x + y)n are 135, 30 and 103, respectively, then 6 (n3 + x2 + y ) is equal to .

Ques. 6JEE Main 2024, 5 April (Morning Shift)

If the constant term in the expansion of (1 + 2x - 3x3 ) ( 32x2 - 13x )9 is p, then 108p is equal to .

Ques. 7JEE Main 2024, 5 April (Evening Shift)

If the constant term in the expansion of ( [5]3x + 2x [3]5 )12, x 0, is 28 [5]3, then 25 is equal to :

  1. 724
  2. 742
  3. 693
  4. 639
Ques. 8JEE Main 2024, 4 April (Morning Shift)

The sum of all rational terms in the expansion of (2 15+5 13 )15 is equal to :

  1. 633
  2. 6131
  3. 3133
  4. 931
Ques. 9JEE Main 2024, 4 April (Evening Shift)

If the coefficients of x4, x5 and x6 in the expansion of (1 + x)n are in the arithmetic progression, then the maximum value of n is:

  1. 28
  2. 21
  3. 7
  4. 14
Ques. 10JEE Main 2024, 1 February (Morning Shift)

If the coefficient of x30 in the expansion of (1+ 1x )6 (1+x2 )7 (1-x3 )8; x 0 is , then | | equals .

Ques. 11JEE Main 2024, 1 February (Evening Shift)

Let m and n be the coefficients of seventh and thirteenth terms respectively in the expansion of ( 13x 13+ 12x 23 )18. Then ( nm ) 13 is :

  1. 19
  2. 14
  3. 49
  4. 94

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