Since n is an even integer. Therefore, n = 2m, m ∈ N.

Also, k = ⇒ k = 3m.

∴ 3n C 2r–1

= 6m C 2r–1

= 6m C 1 – 6m C 3 (3) + 6m C 5 (3) 2 – 6m C 7 (3) 3 + ….. + (–3) 3m–1 6m C 6m–1

Now, (1 + i ) 6m =

⇒ 2 6m

= 6m C 0 + 6m C 1 (i ) + 6m C 2 (i ) 2 + ….

….+ 6m C 3 (i ) 3 + … + 6m C 6m (i ) 6m

⇒ 2 6m (cos 2m π + i sin 2m π )

= { 6m C 0 – 6m C 2 3 + 6m C 4 3 2 – ….}

+i { 6m C 1 – 6m C 3 (3) + 6m C 5 (3) 4 – ….}

[Using de moivre’s theorem on L.H.S.]

Equating imaginary parts on both sides, we get

2 6m sin 2m π = { 6m C 1 – 6m C 3 (3) + 6m C 5 (3) 4 – ….}

⇒ 6m C 1 – 6m C 3 (3) + 6m C 5 (3) 4 – …. = 0 [  sin 2m π = 0]

⇒ 6m C 2r–1 = 0

⇒ 3n C 2r – 1 = 0, where n = 2m and k = .

Hence is correct answer.