For the first sum,
log5k3k2=kk2log53=klog53, so
k=1∑20klog53=220⋅21log53=210log53.
For the second sum,
log9k25k=log925=log35, independent of
k, so the sum is
100log35.
Since
log53⋅log35=1, the product is
210⋅100=21,000.
Thus, the correct answer is
E.