For odd values of r, the value will be negative, and for even, it will be positive.
So, break it in two series :
(0+2)(0+1) + (2+2)(2+1) + (4+2)(4+1) + ... +(20+2)(20+1) =
$\sum_{t=0}^{10}(2t+2)(2t+1) = \sum_{t=0}^{10}(4t^{2}+6t + 2)$ = 1890
And :
-[(1+2)(1+1) +(3+2)(3+1) + (5+2)(5+1) +...+ (19+2)(19+1)] =
$\sum_{t=1}^{10}(2t-1+2)(2t-1+1) = \sum_{t=1}^{10}(2t)(2t+1) = \sum_{t=1}^{10}(4t^{2}+2t)$ = 1644
Total sum = 1890 - 1644 = 246