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The first three terms in the expansion o...

The first three terms in the expansion of `(1+a x)^n(n!=0)` are `1,6x and 16 x^2dot` Then find the value of `a and ndot`

Text Solution

Verified by Experts

The correct Answer is:
`a=2//3, n = 9`
`T_(1) = .^(n)C_(0)=1`
`T_(2) = .^(n)C_(1) ax = 6x " "(1)`
`T_(3) = .^(n)C_(2) (ax)^(2) = 16x^(2) " "(2)`
From (2),
`na = 6 " " (3)`
From (3),
`(n(n-1))/(2)a^(3) = 16 " "(4)`
Eliminating a from (3) and (4),
`(n-1)/(2n) = 4/9` or `n = 9`
From (3),
`a = 2//3`
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Knowledge Check

  • The first three terms in the expansion of (1+ ax) ^(n) are 1, 6x and 16x^(2), then the values of a and n are -

    A
    `a=2, n=9`
    B
    `a=2, n=3`
    C
    `a=3/2, n=6`
    D
    `a= 2/3, n=9`

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

    A
    `2 ^nC_n`
    B
    `(1.3.5…..(2n-1))/(n!)`
    C
    2.6….(4n-2)
    D
    `((2n)!)/((n!)(n!))`

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