Final answer:
A combinatorial proof of the identity 1 + 2 + 3 + ... + n = (n+1)C2 involves counting pairs from a set of n+1 elements in two ways: by summing up choices for the second element, and by using the choose function for selecting 2-element subsets.
Step-by-step explanation:
To provide a combinatorial proof for the identity 1 + 2 + 3 + ... + n = (n+1)C2, consider a set of n+1 elements. We want to count the number of ways to select pairs from this set. When we choose a pair, there is always a smallest element in the pair, and this element can be any one of the n elements (not counting the largest element, which cannot be the smallest in a pair by definition).
For each choice of the smallest element, there are as many choices for the second element of the pair as the value of the smallest element itself: if the smallest element is 1, there's 1 choice for the second element; if it's 2, there are 2 choices; and so on, up to n-1 choices when the smallest element is n. Summing these up gives us the left-hand side of the identity: 1 + 2 + ... + n.
On the other hand, the right-hand side of the identity (n+1)C2 counts the number of 2-element subsets from an (n+1)-element set, which corresponds to the number of pairs we can form. This is because for each 2-combination, we're simply selecting a pair from our set without regard to order. The combinatorial expression (n+1)C2 expands to the fraction \((n+1)n/2\), which simplifies to (n(n+1))/2, matching the sum of the series 1 + 2 + ... + n.
Hence, by two different methods of counting the same thing (the pairs formed from our set), we have shown that 1 + 2 + 3 + ... + n equals (n+1)C2, giving us a combinatorial proof of the identity.