The numbers an = 6n - 5n for n = 1, 2, 3, . . . when divided

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 Multiple Choice QuestionsMultiple Choice Questions

81.

If x is numerically so small so that x and higher powers of x can be neglected, then 1 + 2x33232 +5x- 15 is approximately equal to

  • 32 + 31x64

  • 31 + 32x64

  • 31 - 32x64

  • 1 - 2 x64


82.

For x < 1, the constant term in the expansion of 1x - 12x - 2 is

  • 2

  • 1

  • 0

  • - 12


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83.

The numbers an = 6n - 5n for n = 1, 2, 3, . . . when divided by 25 leave the remainder

  • 9

  • 7

  • 3

  • 1


D.

1

Given an = 6n - 5n, n = 1, 2, 3, . . . We take; 6n = 1 + 5nExpand with binomial expansion6n = C0n + C1n 5 + C2n 52 + C3n 53 + . . .6n = 1 + n 5 + C2n 25 + C3n 53 + . . 6n - 5n = 1 + 25C2n + C3n . 5 + . . .6n - 5n = 1 + 25kwhere k = positive integer.Hence, an = 6n - 5n  divided by 25 and leave the remainder = 1


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84.

For x < 15, the coefficient of x3 in the expansion of 11 - 5x1 - 4x  is

  • 369

  • 370

  • 371

  • 372


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85.

If the coefficients of rth and (r + 1)th terms in the expansion of (3 + 7x)29 are equal, then r is equal to

  • 14

  • 15

  • 18

  • 21


86.

If ab  0 and the sum of the coefficients of x7and x4 in the expansion of x2a - bx11 is 0, then

  • a = b

  • a + b = 0

  • ab = - 1

  • ab = 1


87.

Suppose X follows a binomial distribution with parameters n and p, where 0 < p < 1. If PX = rPX = n - r is independent of n for every r, then p is equal to

  • 12

  • 13

  • 14

  • 18


88.

A fair coin is tossed 100 times. The probability of getting tails an odd number of times is

  • 12

  • 14

  • 18

  • 38


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89.

The value of x  R  loglog1.61 - x2 - 0.62561 + x  R is

  • - , - 1  7, 0

  • (- 1, 5)

  • (1, 7)

  • (- 1, 7)


90.

12 . 3 + 14 . 5 + 16 . 7 + 18 . 9 + ...  = ?

  • log2e

  • loge2

  • log2e

  • e - 1


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