Answer
See below.
Work Step by Step
1. Test for $n=1$, we have $LHS=1\cdot2=2$ and $RHS=\frac{1}{3}(1)(1+1)(4(1)-1)=2$, thus it is true.
2. Assume the formula is true for $n=k$, we have
$1\cdot2+3\cdot4+5\cdot6+...+(2k-1)(2k)=\frac{1}{3}(k)(k+1)(4k-1)$
3. For $n=k+1$, we have $LHS=1\cdot2+3\cdot4+5\cdot6+...+(2k-1)(2k)+(2k+1)(2k+2)=\frac{1}{3}(k)(k+1)(4k-1)+(2k+1)(2k+2)=\frac{1}{3}(k+1)(4k^2-k+12k+6)=\frac{1}{3}(k+1)(4k^2+11k+6)=\frac{1}{3}(k+1)(k+2)(4k+3)=RHS$
4. We conclude that the formula is true for any $n$.
