The sum of series11 × 2C025 + 12 &

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 Multiple Choice QuestionsMultiple Choice Questions

61.

The value of

100011 × 2 + 12 ×3 + 13 ×4 + ... + 1999 × 1000

  • 1000

  • 999

  • 1001

  • 1999


62.

Six positive numbers are in GP, such that their product is 1000. If the fourth term is 1, then the last term is

  • 1000

  • 100

  • 1100

  • 11000


63.

Five numbers are in AP with common difference  0. If the 1st, 3rd and 4th terms are in GP, then

  • the 5th term is always 0.

  • the 1st term is always 0.

  • the middle term is always 0

  • the middle term is always - 2.


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64.

The sum of series

11 × 2C025 + 12 × 3C125 + 13 × 4C225 + ... + 126 × 27C2525

is

  • 227 - 126 × 27

  • 227 - 2826 × 27

  • 12226 + 126 × 27

  • 226 - 152


B.

227 - 2826 × 27

Given series is,

11 × 2C025 + 12 × 3C125 + 13 × 4C225 + ... + 126 × 27C25250x1 + x25dx = 0xC025 + C125x + C225x2 + ... + C2525x25dxOn integrating w.r.t. x, taking limits 0 to x, we get1 + x26260x = C025 + C125 . x22 + C225 . x33 + ... + C25 25. x26260x 1261 + x26 - 126 = C025 + C125 . x22 + C25 25. x2626Again, integrating w.r.t. x, taking limits 0 to 1, we get

126011 + x26 - 1dx= 01C025x + C025 . x22 + ... + C2525x2626dx 1261 + x2727 - x01= C025 . x22 +  C125 . x32 × 3 +  ... + C2525x2726 × 2701 12622727 - 1 - 127 = 12C025 + 12 × 3C125 + ... + 126 × 27C2525 11 × 2 . C025 + 12 × 3C125 + 13 × 4C225 + ... + 126 ×27C2525= 227 - 2826 × 27


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65.

Let f : R  R  be such that f is injective and f(x) f(y) = f(x + y) for all x, y if f(x), f(y) and f(z) are in GP, then x, y and z are in

  • AP always

  • GP always

  • AP depending on the values of x, y and z

  • GP depending on the values of x, y and z


66.

If P = 1 + 12 × 2 + 13 × 22 + ... and Q = 11 × 2 + 13 × 4 + 15 × 6 + ...,

then

  • P = Q

  • 2P =Q

  • P = 2Q

  • P = 4Q


67.

If x = 1 + 12 × 1! + 14 × 2! + 18 × 3! + ... and y = 1 + x21! + x42! + x63! +... Then, the value of logey is

  • e

  • e2

  • 1

  • 1e


68.

The value of the infinite series

12 + 223! + 12 + 22 + 324! + 12 + 22 + 32 +425! + ... is

  • e

  • 5e

  • 5e6 - 12

  • 5e6


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69.

The sum of the series 1 + 12C1n + 13C2n + ... + 1n + 1Cnn is equal to

  • 2n + 1 - 1n + 1

  • 32n - 12n

  • 2n + 1n + 1

  • 2n + 12n


70.

The value of r = 21 + 2 + ... + r - 1r!

  • e

  • 2e

  • e2

  • 3e2


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