Sol. Differentiating the given identity and multiplying by x,

n (1 + x + x 2 ) n–1 (1 + 2x) x = a 1 x + 2a 2 x 2 + 3a 3 x 3 + 4a 4 x 4 + ………….

If ω ≠ 1 is a cube root of unity, we have 1 + ω + ω 2 = 0 and x = 1 gives n3 n = a 1 + 2a 2 + 3a 3 + 4a 4 +……

x = ω gives 0 = a 1 ω +2a 2 ω 2 + 3a 3 + 4a 4 ω + …

x = ω 2 gives 0 = a 1 ω 2 + 2a 2 ω + 3a 3 + 4a 4 ω 2 +…

Adding and dividing by 9, we get

a 3 + 2a 6 + 3a 9 + ………= n 3 n–2 Similarly substituting x = 1, ω , ω

2 in .

n (1 + x + x 2 ) n–1 (1 + 2x) =a 1 + 2a 2 x + 3a 3 x 2 +………. and adding, we get

a 1 + 4a 4 + 7a 7 + ………= n. 3 n–1 It follo0ws that a 1 –a 3 + 4a 4 –2a 6 + 7a 7 –3a 9 + ….. = 2n. 3

n–2