Prove: 
Proof. We prove by induction. If
, then the statement is true since 
. Assume then that the statement is true for some 
. So,
![\[ \sum_{k=1}^m 1 = m \quad \implies \quad \left( \sum_{k=1}^m 1 \right) + 1 = m+1 \quad \implies \quad \sum_{k=1}^{m+1} 1 = m+1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-37b5641cc140b635d4dda3065b84a8f7_l3.png "Rendered by QuickLaTeX.com")
Where the final step follows since the value of 1 for 
is still 1. (Maybe this could confuse since we are summing over the index 
, but the value is independent of . So, really, we are just counting… so for each in the index we add 1; thus, when we have the sum from 
to 
and add 1, it is the same as summing from to 
). Thus, by induction, the statement is true for all 
You can also do this using a telescoping sum, as we are instructed to use the properties derived in Exercise 2 as much as possible.
How do we do this with a telescoping sum when every a_k = 1?
Let 1 be equal to k-(k-1) and use the associative property to cancel out the body of the sum leaving only n-(k-1) as the result which is n.