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Binomial Theorem · Question 19

JEE Main 2025 (Online) 8th April Evening ShiftMCQEasy📈 Topic in 19% papers+4−1
1

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)

JEE Main 2025 (Online) 8th April Evening Shift

Keep practising — Binomial Theorem

2

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

JEE Main 2025 (Online) 8th April Evening Shift
3

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

  1. 4
  2. 6
  3. 3
  4. 1
JEE Main 2025 (Online) 7th April Morning Shift
4

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

JEE Main 2025 (Online) 7th April Evening Shift
5

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
JEE Main 2025 (Online) 4th April Morning Shift
6

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
JEE Main 2025 (Online) 4th April Morning Shift
7

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
JEE Main 2025 (Online) 4th April Evening Shift
8

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

  1. 16923
  2. 18817
  3. 3763
  4. 33845
JEE Main 2025 (Online) 3rd April Morning Shift
9

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

  1. 27
  2. 81
  3. 18
  4. 9
JEE Main 2025 (Online) 3rd April Morning Shift
10

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

JEE Main 2025 (Online) 3rd April Evening Shift
11

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
JEE Main 2025 (Online) 2nd April Morning Shift

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