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Binomial Theorem

JEE Main 2025 (Online) 4th April Morning ShiftMCQMedium📈 Topic in 43% papers+4−1
Ques. 1JEE 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

Answer

(4) 3 2

Solution

In (x + y)2n - 3 there are 2n - 2 coefficients and their sum is 22n - 3 (put x = y = 1), so the arithmetic mean is

22n - 32n - 2 = 16

22n - 3 = 32(n - 1)

Trying integers n 2: n = 5 gives 27 = 128 = 32 4 (the left side grows much faster, so n = 5 is the only solution).

Hence P(2n - 1, n2 - 4n) = P(9, 5), and its distance from x + y - 8 = 0 is

d = |9 + 5 - 8| 12 + 12 = 6 2 = 3 2

Answer: (4)

Keep practising — Binomial Theorem

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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. 7JEE 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. 8JEE 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. 9JEE 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
Ques. 10JEE Main 2025, 29 January (Evening Shift)

The remainder, when 7103 is divided by 23, is equal to:

  1. 9
  2. 6
  3. 14
  4. 17
Ques. 11JEE Main 2025, 28 January (Morning Shift)

If = 1 + r=16(-3)r-1 12C2r-1, then the distance of the point (12, 3) from the line x - 3y + 1 = 0 is .

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