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Binomial Theorem

JEE Main 2026 (Online) 22nd January Evening ShiftMCQMedium📈 Topic in 43% papers+4−1
Ques. 1JEE Main 2026, 22 January (Evening Shift)

Let Cr denote the coefficient of xr in the binomial expansion of (1 + x)n, n N, 0 r n. If Pn = C0 - C1 + 223C2 - 234C3 + + (-2)nn+1Cn, then the value of n=125 1P2n equals

  1. 675
  2. 580
  3. 525
  4. 650

Answer

(1) 675

Solution

General term: Pn = r=0n (-2)rr+1Cr

Crr+1 = 1n+1 n+1Cr+1

Pn = 1n+1 r=0n n+1Cr+1(-2)r = 1-2(n+1) k=1n+1 n+1Ck(-2)k

= (1 - 2)n+1 - 1-2(n+1) = 1 - (-1)n+12(n+1)

For 2n in place of n: P2n = 1 + 12(2n+1) = 12n+1

n=125 1P2n = n=125(2n + 1) = 2 25 262 + 25

= 650 + 25 = 675

Answer: (1)

Keep practising — Binomial Theorem

Ques. 2JEE Main 2026, 21 January (Morning Shift)

If the coefficient of x in the expansion of (ax2 + bx + c)(1 - 2x)26 is -56 and the coefficients of x2 and x3 are both zero, then a + b + c is equal to :

  1. 1483
  2. 1300
  3. 1500
  4. 1403
Ques. 3JEE Main 2026, 21 January (Evening Shift)

If ( 115C0 + 115C1 ) ( 115C1 + 115C2 ) ( 115C12 + 115C13 ) = 1314C0 14C1 14C12, then 30 is equal to .

Ques. 4JEE 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
Ques. 5JEE Main 2025, 8 April (Evening Shift)

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

Ques. 6JEE 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. 7JEE 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. 8JEE 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. 9JEE 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. 10JEE 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. 11JEE 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

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