On the q-Analogue of the Sum of Cubes

A simple q-analogue of the sum of cubes is given. This answers a question posed in this journal by Garrett and Hummel. The sum of cubes and its q-analogues It is well-known that the first n consecutive cubes can be summed in closed form as n ∑ k=1 k = ( n + 1 2 )2 . Recently, Garrett and Hummel discovered the following q-analogue of this result: n ∑ k=1 qk−1 (1− q)(2− qk−1 − q) (1− q)2(1− q2) = [ n + 1 2 ]2 , (1) where [ n k ] = (1− qn−k+1)(1− qn−k+2) · · · (1− q) (1− q)(1− q2) · · · (1− qk) is a q-binomial coefficient. In their paper Garrett and Hummel commiserate the fact that (1) is not as simple as one might have hoped, and ask for a simpler sum of q-cubes. In response to this I propose the identity n ∑ k=1 q2n−2k (1− q)(1− q) (1− q)2(1− q2) = [ n + 1 2 ]2 . (2) ∗Work supported by the Australian Research Council the electronic journal of combinatorics 11 (2004), #R00 1 Proof. Since [ n + 1 2 ]2 − q [ n 2 ]2 = (1− q)(1− q) (1− q)2(1− q2) equation (2) immediately follows by induction on n. The form of (2) should not really come as a surprise in view of the fact that the q-analogue of the sum of squares n ∑ k=1 k = 1 6 n(n + 1)(2n + 1)