What is the asymptotic bound for $1n + 2(n-1) + 3(n-2) + ... + (n-1)2 + n$?
We can lower bound the sum roughly by
$$
\sum_{i=1}^n i(n+1-i) \geq \sum_{i=n/3}^{2n/3} (n/3)^2 \geq (n/3)^3.
$$
This shows that it is $\Omega(n^3)$. Since each summand is at most $n^2$, the sum is also $O(n^3)$.

We can also compute the sum explicitly:
$$
\sum_{i=1}^n i(n+1-i) = (n+1) \sum_{i=1}^n i - \sum_{i=1}^n i^2 = (n+1) \frac{(n+1)n}{2} - \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(n+2)}{6}.
$$
This expression equals the binomial coefficient $\binom{n+2}{3}$, and we can prove this bijectively. The binomial coefficient counts triples $(a,b,c)$ of elements satisfying $0 \leq a &lt; b &lt; c \leq n+1$. Notice that $1 \leq b \leq n$, and this corresponds to summation over $i$. Given $b$, we have $b$ choices for $a$ and $n+1-b$ choices for $c$.