5.1. Calculate: $\frac{1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+2013 \cdot 2014}{(1+2+3+\ldots+2014) \cdot \frac{1}{5}}$. If necessary, round the answer to the nearest hundredths.
Answer: 6710. Solution. Let's calculate the sum in the numerator: $\sum_{k=1}^{n} k(k+1)=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k$ $=\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{3}$. Here, the well-known formula for the sum of squares of natural numbers is used, which can be proven in various ways (these proofs can easily be found in literature).

Thus, the fraction specified in the condition is equal to $\frac{2013 \cdot 2014 \cdot 2015 \cdot 2 \cdot 5}{3 \cdot 2014 \cdot 2015}=\frac{2013 \cdot 10}{3}=6710$.