Hey folks, let's start by noting something important first — the identity you wrote $\binom{2n}{n}=2\binom{n}{2}+n^2$ doesn't hold for most integer values of $n$ (for example, when $n=3$, the left side is $\binom{6}{3}=20$, while the right side is $2*\binom{3}{2}+3^2=6+9=15$, which aren't equal). I suspect there's a small typo here, and you probably meant to prove $\binom{2n}{2}=2\binom{n}{2}+n^2$ — that identity holds true for all positive integers $n$, so let's work through that algebraic proof step by step.

Step 1: Recall the simplified combination formula for choosing 2 elements #

For any integer $k \geq 2$, the combination $\binom{k}{2}$ (which counts the number of ways to choose 2 elements from $k$ total) simplifies to:
$$\binom{k}{2} = \frac{k(k-1)}{2}$$
This comes directly from the general combination formula $\binom{k}{m} = \frac{k!}{m!(k-m)!}$ — when $m=2$, the factorials simplify down to the product of the first two descending integers from $k$, divided by 2.

Step 2: Expand the Right-Hand Side (RHS) #

Let's break down $2\binom{n}{2} + n^2$ piece by piece:

• First calculate $\binom{n}{2}$: $\binom{n}{2} = \frac{n(n-1)}{2}$
• Multiply by 2 to eliminate the denominator: $2*\binom{n}{2} = 2*\frac{n(n-1)}{2} = n(n-1) = n^2 - n$
• Add $n^2$ to the result: $n^2 - n + n^2 = 2n^2 - n$

Step 3: Expand the Left-Hand Side (LHS) #

Now compute $\binom{2n}{2}$ using the simplified formula we recalled earlier:
$$\binom{2n}{2} = \frac{(2n)(2n-1)}{2}$$
Let's simplify this expression:
$$\frac{(2n)(2n-1)}{2} = \frac{4n^2 - 2n}{2} = 2n^2 - n$$

Step 4: Verify equality #

After expanding both sides, we can see that:

• LHS = $2n^2 - n$
• RHS = $2n^2 - n$

Since both sides simplify to the exact same algebraic expression, we've proven that $\binom{2n}{2}=2\binom{n}{2}+n^2$.

If you truly intended to work with $\binom{2n}{n}$ instead, feel free to clarify — that identity would need adjustment, as it doesn't hold for most values of $n$.

内容的提问来源于stack exchange，提问作者user104