2010-09-15, 12:13 AM
Probably the easiest thing to do is to distribute out both sides. You should get that:
SUM[(2r-1)³] from r=1 to k+1 = k²(2k²-1)+(2k+1)³ = 2k^4+8k³+11k²+6k+1
and you should get that:
(k+1)²(2(k+1)²-1) = 2k^4+8k³+11k²+6k+1
and hence the two are equal.
SUM[(2r-1)³] from r=1 to k+1 = k²(2k²-1)+(2k+1)³ = 2k^4+8k³+11k²+6k+1
and you should get that:
(k+1)²(2(k+1)²-1) = 2k^4+8k³+11k²+6k+1
and hence the two are equal.