Q.

The sum of all possible values of n∈N, so that the coefficients of x,x2 and x3 in the expansion of (1+x2)2(1+x)n, are in arithmetic progression, is:           [2026]

1 3  
2 7  
3 9  
4 12  

Ans.

(3)

(x4+2x2+1)(C0nx0+C1nx1+C2nx2+C3nx3+…)

Coefficient of x⇒C1n

Coeff. of x2⇒2+C2n=2+n(n-1)2

Coeff. of x3=2·C1n+C3n

=2n+n(n-1)(n-2)6  (if x≥3)

Now according to question

n+2n+n(n-1)(n-2)6=2[2+n(n-1)2]

3n+n(n-1)(n-2)6=4+n(n-1)

⇒n3-9n2+26n-24=0

⇒n=2,3,4  ⇒n=3,4

Now checking for n=2

Coeff. of x=2Coeff. of x2=3Coeff. of x3=4}⇒are in A.P.

⇒n=2 is also the correct choice

Required sum of values of n

=2+3+4=9

Option (3)