Splitting summations?
In the first sum, you know that k ≥ 0. In the second sum, you know k ≥ n/2. Therefore, adding k in the first sum is ≥ adding 0 in the first sum, and adding k in the second sum is ≥ adding n/2 in the second sum. 

You can adapt this for example to show that the sum of $k^5$ for 1 ≤ k ≤ n is $O(n^6)$: You have n/2 terms each at least $n^5/32$, for a total of $n^6/64 = O(n^6)$. Or a lower bound for n!, where you have a product including n/2 terms each larger than n/2, so n! > $(n/2)^{n/2}$.