For any positive integer m, n (with n ≥ m), let (n/m) = nCm. Prove that (n/m) + (n-1/m) + (n-2/m) +

Q: Wait, why can we just replace $\binom{m}{m}$ with $\binom{m+1}{m+1}$? Isn't that changing the value?

That is a brilliant observation! We aren't changing the value, we are changing the 'form.' Since $\binom{m}{m} = 1$ and $\binom{m+1}{m+1} = 1$, they are mathematically identical. We perform this substitution to align the indices so that Pascal's Identity can 'see' the pattern and begin the chain reaction. It is like changing a dollar bill into four quarters—the value is the same, but the utility for the transaction is much higher!

Q: The second part has coefficients like $1, 2, 3...$ How do I handle that? It looks like a mess.

It looks intimidating, but think of it as a 'sum of sums.' If you have $2\binom{n-1}{m}$, that is just $\binom{n-1}{m} + \binom{n-1}{m}$. By breaking the coefficients into separate rows, you reveal that the complex sum is actually just a stack of Hockey-stick identities waiting to be solved. You are not solving one big problem; you are solving a series of smaller, identical problems.

Q: Can I prove this using induction instead?

Absolutely! Induction is a powerful tool here. You would prove the base case for $n=m$, then assume it is true for $n=k$ and show it holds for $n=k+1$. However, the telescoping method we used is often more intuitive because it shows *why* the terms collapse, rather than just verifying the result. Both are valid, but the telescoping method builds better intuition for how these numbers interact.

Question

For any positive integer $m, n$ (with $n \ge m$), let $\binom{n}{m} = ^nC_m$. Prove that $\binom{n}{m} + \binom{n-1}{m} + \binom{n-2}{m} + .... + \binom{m}{m} = \binom{n+1}{m+1}$. Hence or otherwise, prove that $\binom{n}{m} + 2\binom{n-1}{m} + 3\binom{n-2}{m} + .... + (n - m + 1)\binom{m}{m} = \binom{n+2}{m+2}$.

Sigma Percentile · Accepted Answer

### The Elegance of the Hockey-stick Identity: A Journey Through Pascal's Triangle

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering a hidden symmetry in the heart of combinatorics. We are going to explore the **Hockey-stick Identity**.

It sounds like a sports term, but it is actually one of the most beautiful visual proofs in all of mathematics. Imagine you are standing before Pascal's Triangle, that infinite pyramid of numbers where each entry is the sum of the two above it.

If you pick a diagonal line starting from the edge and sum the numbers along that diagonal, you will find a result that bends like a hockey stick, landing on the number below and to the right of the last term in your sum. Let us prove this mystery.

### Phase 1: The Master Key

To unlock this, we need our master key: **Pascal's Identity**. It states that:

$$\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$$

Physically, this tells us that if you take two adjacent numbers in a row of Pascal's Triangle, their sum is the number directly below them. This is the engine of our proof. We want to prove that:

$$\sum_{r=m}^n \binom{r}{m} = \binom{n+1}{m+1}$$

Let us write out the sum:

$$S = \binom{m}{m} + \binom{m+1}{m} + \binom{m+2}{m} + \dots + \binom{n}{m}$$

Here is the trick: we need to start the chain reaction. We know that $\binom{m}{m} = 1$ and $\binom{m+1}{m+1} = 1$. So, let us replace the first term $\binom{m}{m}$ with $\binom{m+1}{m+1}$. Now our sum looks like this:

$$\binom{m+1}{m+1} + \binom{m+1}{m} + \binom{m+2}{m} + \dots + \binom{n}{m}$$

### Phase 2: The Domino Effect

Look at the first two terms: $\binom{m+1}{m+1} + \binom{m+1}{m}$. The upper indices are the same ($m+1$), and the lower indices are consecutive ($m+1$ and $m$). Pascal's Identity screams at us to combine them, collapsing into $\binom{m+2}{m+1}$.

Now, our sum has become:

$$\binom{m+2}{m+1} + \binom{m+2}{m} + \dots + \binom{n}{m}$$

Do you see it? The new term $\binom{m+2}{m+1}$ is now ready to combine with the next term in the sequence, $\binom{m+2}{m}$. They merge to form $\binom{m+3}{m+1}$.

This is a beautiful, cascading collapse. Like a row of dominoes, each term combines with the next, pushing the sum forward until we reach the very end. The final result is $\binom{n+1}{m+1}$. We have proved the **Hockey-stick Identity**!

### Phase 3: The Matrix of Sums

Now, let us tackle the second part of the problem. We are asked to prove:

$$\binom{n}{m} + 2\binom{n-1}{m} + 3\binom{n-2}{m} + \dots + (n - m + 1)\binom{m}{m} = \binom{n+2}{m+2}$$

At first glance, those coefficients ($1, 2, 3, \dots$) look terrifying. But do not panic. Let us use the **Row Expansion** strategy. If you have a coefficient of $2$, it just means you have two copies of that term.

Let us arrange these into a grid:

Row 1: $\binom{n}{m} + \binom{n-1}{m} + \dots + \binom{m}{m}$

Row 2: $\binom{n-1}{m} + \binom{n-2}{m} + \dots + \binom{m}{m}$

...

Row $(n-m+1)$: $\binom{m}{m}$

When you sum these rows vertically, you get exactly the expression in the problem. But look at the rows horizontally! Each row is just the Hockey-stick Identity we just proved.

### Phase 4: The Final Victory

Let us evaluate each row using our result from Phase 2:

The sum of Row 1 is $\binom{n+1}{m+1}$.

The sum of Row 2 is $\binom{n}{m+1}$.

The sum of the last row is $\binom{m+1}{m+1}$.

Now, our total sum is simply the sum of these results:

$$\binom{n+1}{m+1} + \binom{n}{m+1} + \dots + \binom{m+1}{m+1}$$

Wait! Look at this new expression. It is another Hockey-stick series! The lower index is now $m+1$, and the upper index runs from $m+1$ to $n+1$.

Applying the identity one last time, the entire sum collapses into **$\binom{n+2}{m+2}$**. We have arrived at the destination. You have successfully navigated the Hockey-stick Identity, mastering a fundamental tool of combinatorics.

Visualized Solution

The Hockey-stick Identity

Pascal's Identity

The First Substitution: $(m m) = (m + 1 m + 1)$

The First Combination

Collapsing the Series

Analyzing Part Two

Expanding the Sum into Rows

Applying Part One to Each Row

The Final Summation

The Sigma Insight: Properties of Binomial Coefficients

The Elegance of the Hockey-stick Identity

Phase 1

Phase 2

Phase 3

Phase 4

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