We have,

(1 + x + x 2 ) n = a 0 + a 1 x + a 2 x 2 + ….. + a 2n x 2n …(1)

and (x – 1) n = n C 0 x n – n C 1 x n–1 + n C 2 x n–2 –….. + (–1) n n C n x n …(2)

Multiplying (1) and (2), we get (1 + x + x 2 ) n (x – 1) n

= (a 0 + a 1 x + a 2 x 2 + ….. + a 2n x 2n ) × { n C 0 x n – n C 1 x n–1 + n C 2 x n–2 –….+(–1) n n C n }

⇒ (x 3 – 1) n = (a 0 + a 1 x + a 2 x

2 + …. + a 2n x 2n ) × { n C 0 x n – n C 1 x n–1 +….

+(–1) n n C n } ….(3)

Now, coefficient of x n on RHS of (3)

= a 0 n C 0 – a 1 n C 1 + a 2 n C 2 – ….. + (–1) n a n n C n

LHS of (3)

= (x 3 –1) n

= (–1) n (1 – x 3 ) n

= (–1) n (– x 3 ) r

= (–1) n n C r x 3r ….(4)

Clearly, if n is not a multiple of 3, then x n does not occur in (4)

∴ (Coefficient of x n in LHS) = 0,

when n is not a multiple of 3. If n is a multiple of 3 i.e. if

n = 3m, then

= (–1) 3m (–1) m 3m C m

= 3m C m = n C n/3 [  n = 3m]

Thus, equation the coefficients of x n on both sides, we get

a 0 n C 0 – a 1 n C 1 + a 2 n C 2 – a 3 n C 3 + ….. + (–1) n a n n C n

=

Hence is correct answer.