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Binomial Theorem

JEE Main 2025 (Online) 7th April Morning ShiftMCQEasy📈 Topic in 10% papers+4−1
Ques. 1JEE Main 2025, 7 April (Morning Shift)

The remainder when ((64)(64) )(64) is divided by 7 is equal to

  1. 4
  2. 6
  3. 3
  4. 1

Answer

(4) 1

Solution

64 = 63 + 1 = 7 9 + 1, so 64 1 7, and by the binomial expansion

(64)64 = (1 + 63)64 = 1 + 7k

((64)64 )64 = (1 + 7k)64 = 1 + 7m

remainder = 1

Answer: (4)

Keep practising — Binomial Theorem

Ques. 2JEE Main 2025, 7 April (Evening Shift)

The sum of the series 2 1 20C4 - 3 2 20C5 + 4 3 20C6 - 5 4 20C7 + + 18 17 20C20, is equal to .

Ques. 3JEE Main 2025, 4 April (Morning Shift)

For an integer n 2, if the arithmetic mean of all coefficients in the binomial expansion of (x + y)2n - 3 is 16, then the distance of the point P (2n - 1, n2 - 4n ) from the line x + y = 8 is

  1. 2
  2. 2 2
  3. 5 2
  4. 3 2
Ques. 4JEE Main 2025, 4 April (Morning Shift)

In the expansion of ( [3]2 + 1 [3]3 )n, n N, if the ratio of 15th term from the beginning to the 15th term from the end is 16, then the value of nC3 is

  1. 4960
  2. 2300
  3. 1040
  4. 4060
Ques. 5JEE Main 2025, 4 April (Evening Shift)

If 12 (15C1 ) + 22 (15C2 ) + 32 (15C3 ) + + 152 (15C15 ) = 2m 3n 5k, where m, n, k N, then m + n + k is equal to :

  1. 20
  2. 19
  3. 18
  4. 21
Ques. 6JEE Main 2025, 3 April (Morning Shift)

The sum of all rational terms in the expansion of (2+ 3)8 is :

  1. 16923
  2. 18817
  3. 3763
  4. 33845
Ques. 7JEE Main 2025, 3 April (Morning Shift)

If r=19 ( r+32r ) 9Cr = ( 32 )9 - , , N, then ( + )2 is equal to

  1. 27
  2. 81
  3. 18
  4. 9
Ques. 8JEE Main 2025, 3 April (Evening Shift)

Let (1 + x + x2 )10 = a0 + a1x + a2x2 + + a20x20. If (a1 + a3 + a5 + + a19) - 11a2 = 121k, then k is equal to .

Ques. 9JEE Main 2025, 2 April (Morning Shift)

The term independent of x in the expansion of ( (x + 1)(x2/3 + 1 - x1/3) - (x - 1)(x - x1/2) )10, x > 1, is :

  1. 240
  2. 120
  3. 150
  4. 210
Ques. 10JEE Main 2025, 2 April (Evening Shift)

If r=010 ( 10r+1 - 110r ) 11Cr+1 = 11 - 11111010, then is equal to :

  1. 11
  2. 20
  3. 24
  4. 15
Ques. 11JEE Main 2025, 29 January (Morning Shift)

The least value of n for which the number of integral terms in the Binomial expansion of ( [3]7 + [12]11)n is 183, is :

  1. 2184
  2. 2172
  3. 2196
  4. 2148

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