Let `(1+x^2)^2(1+x)^n=sum_(k=0)^(n+4)a_k x^k`. If `a_1`, `a_2` and `a_3` are in arithmetic progression, then the possible value/values of `n` is/are a. 5 b. 4 c. `3` d. `2`

Explanation

We have,
`(1+ x^(2))^(2) (1 + x)^(n) = (1 + 2x^(2) + x^(4))`
` xx(""^(n)C_(0) + ""^(n)C_(1)x + ""^(n)C_(2) x^(2) + ""^(n)C_(3) x^(3) + …)`
` = a_(0) + a_(1) x + a_(2) x^(2) + a_(3) x^(3) + …` [say]
Now , comparing the coefficients of ` x,x^(2) "and " x^(3) ` , we get
` a_(1) = ""^(n)C_(1),A_(2) = 2*""^(n)C_(0) + ""^(n)C_(2),A_(3) = 2 * ""^(n)C_(1) + ""^(n)C_(3)` ....(i)
In ` a_(1) n ge , "in" a_(2), n ge 2 " and in " a_(3) , n ge 3 `
` therefore n ge 3 ` ...(ii)
From Eq.(i) ,
`a_(1) = n,a_(2) = 2+ (n(n-1))/(1*2) = (n^(2) -n +4)/(2)`
and `a_(3) = 2n+ (n(n-1)(n-2))/(1*2*3) =(n^(3) - 3n^(2) + 14n)/(6)`
Since , ` a_(1) , a_(2) ,a_(3) ` are in AP .ltbr. Therefore , ` 2a_(2) = a_(1) + a_(3)`
` rArr n^(2) - n + 4 = n+ (n^(3) - 3n^(2) + 14n)/(6)`
` rArr n^(3) - 9n^(2) + 26n - 24= 0 `
or `(n-2)(n-3)(n-4) = 0 `
` therefore n = 2,3,4 `
Hence , n = 3,4 [from Eq.(ii)]