Is there an easy way to find the closed form of
$$\scriptsize\sum_{r=1}^n (r+\tfrac {2010}{10})(r+2011)(r+2012)(r+2013)(r+2014)(r+2015)(r+2016)(r+2017)(r+2018)$$
without first knowing the answer?
The answer is $$\scriptsize{\frac 1{10}n}(n+2011)(n+2012)(n+2013)(n+2014)(n+2015)(n+2016)(n+2017)(n+2018)(n+2019)$$
Let $p(n)=(n+\tfrac {2010}{10})\prod_{i=2011}^{2018}(n+i)$ and $P(n)=\frac {1}{10}n\prod_{i=2011}^{2019}(n+i)$.
We would like to show that $P(n)=\sum_{r=1}^n p(n)$.
Claim: $P(n)-P(n-1)=p(n)$
Proof: Observe that
\begin{align*} P(n)-P(n-1)&=\frac{1}{10}\left(n(n+2019)-(n-1)(n+2010)\right)\prod_{i=2011}^{2018} (n+1)\\ &=\frac{1}{10}(10n+2010)\prod_{i=2011}^{2018}(n+i)\\ &=p(n) \end{align*}
It follows that
\begin{align*} \sum_{r=1}^n p(r)&=\sum_{r=1}^n \left(P(r)-P(r-1)\right)\\ &=\sum_{r=1}^n P(r)-\sum_{r=0}^{n-1} P(r)\\ &=P(n)-P(0)\\ &=P(n) \end{align*}
As desired $\blacksquare$
