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Binomial Theorem

JEE Main 2026 (Online) 28th January Evening ShiftMCQMedium📈 Topic in 43% papers+4−1
Ques. 1JEE Main 2026, 28 January (Evening Shift)

The sum of the coefficients of x499 and x500 in (1 + x)1000 + x(1 + x)999 + x2(1 + x)998 + + x1000 is :

  1. 1002C501
  2. 1001C501
  3. 1000C501
  4. 1002C500

Answer

(4) 1002C500

Solution

S = k=01000 xk(1 + x)1000-k is a G.P. with first term (1 + x)1000 and ratio x1 + x

S = (1 + x)1000 1 - ( x1 + x )10011 - x1 + x

= (1 + x)1001 - x1001

Coefficient of x499 = 1001C499; coefficient of x500 = 1001C500

Sum = 1001C499 + 1001C500 = 1002C500 ( nCr-1 + nCr = n+1Cr)

Answer: (4)

Keep practising — Binomial Theorem

Ques. 2JEE Main 2026, 24 January (Morning Shift)

Let S = 125! + 13! 23! + 15! 21! + up to 13 terms. If 13 S = 2kn!, k N, then n + k is equal to

  1. 50
  2. 52
  3. 49
  4. 51
Ques. 3JEE Main 2026, 23 January (Morning Shift)

The sum of all possible values of n N, so that the coefficients of x, x2 and x3 in the expansion of (1 + x2)2(1 + x)n, are in arithmetic progression is :

  1. 12
  2. 9
  3. 3
  4. 7
Ques. 4JEE Main 2026, 23 January (Morning Shift)

The value of 100C5051 + 100C5152 + + 100C100101 is:

  1. 2101101
  2. 2100101
  3. 2100100
  4. 2101100
Ques. 5JEE Main 2026, 22 January (Morning Shift)

The coefficient of x48 in (1 + x) + 2(1 + x)2 + 3(1 + x)3 + + 100(1 + x)100 is equal to

  1. 100 100C49 - 100C48
  2. 100 101C49 - 101C50
  3. 100C50 + 101C49
  4. 100 100C49 - 100C50
Ques. 6JEE 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
Ques. 7JEE 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. 8JEE Main 2026, 21 January (Evening Shift)

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

Ques. 9JEE 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. 10JEE Main 2025, 8 April (Evening Shift)

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

Ques. 11JEE 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

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