Show that $\sum_{k=0}^n \binom{k}{i} = \binom{n+1}{i+1}$ for all integers $n\geq 0$ and $i\geq 0$. Again, use a combinatorial argument, where you show that both sides count the same thing.


We are tasked with proving the identity:
\[
\sum_{k=0}^n \binom{k}{i} = \binom{n+1}{i+1}
\]
using a combinatorial argument. Let's break down both sides and interpret them in terms of combinatorics.

### Left-hand side: $\sum_{k=0}^n \binom{k}{i}$

The term $\binom{k}{i}$ represents the number of ways to choose $i$ elements from a set of size $k$. The sum $\sum_{k=0}^n \binom{k}{i}$ counts how many ways we can choose $i$ elements from a set whose size varies from 0 to $n$. So, this sum is considering different choices of $i$-element subsets from sets of sizes $0, 1, 2, \dots, n$.

To understand the total meaning of this, let's imagine we have a set of $n+1$ elements labeled as $ \{0, 1, 2, \dots, n\}$. We are interested in counting how many ways we can choose $i+1$ elements from this set, but we will think about this selection process in a slightly recursive way.

### Right-hand side: $\binom{n+1}{i+1}$

The term $\binom{n+1}{i+1}$ counts the number of ways to choose $i+1$ elements from a set of size $n+1$. This gives the total number of ways to select $i+1$ elements from the entire set $\{0, 1, 2, \dots, n\}$.

### Combinatorial argument:

Now, let's explain how the sum on the left-hand side counts the same thing as the binomial coefficient on the right-hand side.

- Consider the set $ \{0, 1, 2, \dots, n\} $, which has $n+1$ elements.
- We want to count how many ways we can choose $i+1$ elements from this set.
- Break this counting process into cases based on the largest element in the subset of size $i+1$.
  
Let the largest element in the chosen subset be $k$, where $k$ ranges from 0 to $n$. If $k = k$, then the remaining $i$ elements must be chosen from the set $\{0, 1, 2, \dots, k-1\}$, which has $k$ elements.

- The number of ways to choose $i$ elements from the set $\{0, 1, 2, \dots, k-1\}$ is $\binom{k}{i}$.
- Therefore, for each $k$, the number of ways to form an $i+1$-element subset where the largest element is $k$ is $\binom{k}{i}$.

To count all the possible subsets, we sum over all values of $k$, from $k = 0$ to $k = n$. This gives the sum:
\[
\sum_{k=0}^n \binom{k}{i}.
\]

But, by this reasoning, we know that the total number of ways to choose $i+1$ elements from a set of size $n+1$ is $\binom{n+1}{i+1}$.

### Conclusion:

The sum on the left-hand side counts the number of ways to choose $i+1$ elements from a set of size $n+1$ by considering the largest element in the subset. The right-hand side directly counts the same thing. Therefore, both sides of the equation count the same thing, and we conclude that:
\[
\sum_{k=0}^n \binom{k}{i} = \binom{n+1}{i+1}.
\]

Do you need more details or have any questions?

Here are 5 related questions:
1. Can you explain how Pascal's identity is related to this result?
2. How would you derive this identity algebraically rather than combinatorially?
3. How does this sum change if we consider different limits for $k$?
4. Can you generalize this result for other binomial sum identities?
5. What happens when $i > n$ in the original sum?

**Tip:** When counting subsets, it is often helpful to break them down into cases by the largest or smallest element.

Show that $\sum_{k=0}^n \binom{k}{i} = \binom{n+1}{i+1}$ for all integers $n \geq 0$ and $i \geq 0$. Use a combinatorial argument to show that both sides count the same thing.

Discover how to prove the binomial sum identity $\sum_{k=0}^n \binom{k}{i} = \binom{n+1}{i+1}$ using a combinatorial argument. This step-by-step explanation covers binomial coefficients, summation, and key theorems like Pascal's Rule. Suitable for high school students in Grades 10-12.

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