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Let (1 + x^(2))^(2) (1 + x)^(n) = a(0) ...

Let ` (1 + x^(2))^(2) (1 + x)^(n) = a_(0) + a_(1) x + a_(2) x^(2) + …` if
` a_(0),a_(1) " and " a_(2)` are in A.P , the value of n is

A

2

B

3

C

4

D

7

Text Solution

Verified by Experts

The correct Answer is:
b,c
We have ,
Coefficient of ` x , x^(2) " and " x^(3) " in " (1 + x^(2))^(2) (1 + x)^(n)`
i.e. , values of ` a_(1) , a_(2) " and " a_(3) " in " (1 + 2x^(2) + x^(4) ) (1 + x)^(n)`
` rArr a_(1) = ""^(n)C_(1) , a_(2) = ""^(n)C_(2) + 2 " and " a_(3) = ""^(n)C_(3) + 2 ""^(n)C_(1)`
According to the question ,
` 2a_(2) = a_(1) + a_(3)`
`rArr 2(""^(n)C_(2)) = ""^(n)C_(1) + (""^(n)C_(3) + 2^(n)C_(1))`
` rArr [2(n(n-1))/(2) ] + 4 = 3n + (n(n-1)(n-2))/(6)`
` rArr n^(3) - 9n^(2) = 26n = 24 = 0 `
` rArr (n - 2) (n^(2) - 7n + 12) = 0 `
` rArr (n-2)(n-3) (m-4)= 0 `
` rArr n = 2,3,4`
Hence , n == 3,4,(` n ne 2 because ""^(n)C_(3)` is not defined )
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