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If the middle term in the expansion of `(x^2+1//x)^n` is `924 x^6,` then find the value of `ndot`

A

8

B

10

C

12

D

20

Text Solution

Verified by Experts

The correct Answer is:
C
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Knowledge Check

  • The middle term in the expansion of (x^2+1/x^2+2)^n is

    A
    `(2n!)/(n!n!)`
    B
    `(2n!)/(n!)`
    C
    `(n!)/(n-1)`
    D
    None of these

  • The middle term in the expansion of (1+x)^(2n) is

    A
    `""^(2n)C_(n) x^(n)`
    B
    `""^(2n)C_(n+1) x^(n+1)`
    C
    `""^(2n)C_(n-1)x^(n-1)`
    D
    `(1.3.5…(2n-1))/(n!) 2^(n)x^(n)`

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