Let x denote greatest integer less than or equal to x . If for n ∈ N , 1 - x + x 3 n = ∑ j = 0 3 n a j x j , then ∑ j = 0 3 n 2 a 2 j + 4 ∑ j = 0 3 n - 1 2 a 2 j + 1 is equal to :

Step-by-step solution

1 - x + x 3 n = ∑ j = 0 3 n a j x j ⇒ 1 - x + x 3 n = a 0 + a 1 x + a 2 x 2 … … + a 3 n x 3 n       . . . i Then, ∑ j = 0 3 n 2 a 2 j = a 0 + a 2 + a 4 … … ∑ j = 0 3 n - 1 2 a 2 j + 1 = a 1 + a 3 + a 5 … … Put x = 1 in i , then we get 1 = a 0 + a 1 + a 2 + a 3 … … … + a 3 n . . . ( A ) Put x = - 1 in i , then we get 1 = a 0 - a 1 + a 2 - a 3 … … … + - 1 3 n a 3 n . . . ( B ) Solving ( A ) and ( B ) a 0 + a