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Binomial Theorem

JEE Main 2025 (Online) 8th April Evening ShiftMCQEasy📈 Topic in 21% papers+4−1
Ques. 1JEE Main 2025, 8 April (Evening Shift)

The number of integral terms in the expansion of (5 12 + 7 18 )1016 is:

  1. 127
  2. 128
  3. 130
  4. 129

Answer

(2) 128

Solution

Tr+1 = 1016Cr (5 12 )1016 - r (7 18 )r = 1016Cr 5 1016 - r2 7 r8

The term is an integer when r8 and 1016 - r2 are both whole numbers.

r a multiple of 8 r is even 1016 - r is even (1016 is even)

r = 0, 8, 16, , 1016

Number of values = 10168 + 1 = 127 + 1 = 128

Answer: (2)

Keep practising — Binomial Theorem

Ques. 2JEE Main 2025, 8 April (Evening Shift)

The product of the last two digits of (1919)1919 is

Ques. 3JEE 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
Ques. 4JEE 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. 5JEE 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. 6JEE 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. 7JEE 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. 8JEE 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. 9JEE 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. 10JEE 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. 11JEE 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

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