If ab ≠ 0 and the sum of&nb

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 Multiple Choice QuestionsMultiple Choice Questions

251.

If ak is the coefficient of xk in the expansion of 1 + x + x2n for k = 0, 1, 2, . . , 2n, then

  • - a0

  • 3n

  • n 3n + 1

  • n 3n


252.

The coefficient of xk in the expansion of 1 - 2x - x2e- x is

  • 1 - k - k2k!

  • k2 + 1k!

  • 1 - kk!

  • 1k!


253.

The coefficient of x24 in the expansion of (1 + x2)12(1 + x12)(1 + x24) is

  • C612

  • C612 + 2

  • C612 + 4

  • C612 +6


254.

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


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255.

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

  • 2

  • 1

  • 0

  • - 12


256.

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

  • 9

  • 7

  • 3

  • 1


257.

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

  • 369

  • 370

  • 371

  • 372


258.

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


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259.

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


D.

ab = 1

Given expansion is x2a - bx11 The general term isTr + 1 = Cr11 x22 - 3r- br1a11 - r For coefficient x7, put 22 - 3r = 7 3r = 15   r = 5and for coefficient of x4, put 22 - 3r = 4 3r = 18  r =6  T6 = C6111a6- b6According to given condition,T6 +T7 = 0 C5111a5- b51a - b = 0 1a - b = 0        ab = 1


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260.

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


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