Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$

Question

Problem : Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$

My Approach : I would assume that we can prove by induction.

Base case $n=2$.

$$=2{2 \choose 2}+ {2 \choose 1}= (2\cdot 1)+2 =4$$ $$n^2 =2^2 = 4.$$

Assume for $n\ge 2$, $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$ for $n \le k$.

Let $n=k+1$

$$2{k+1 \choose 2}+{k+1 \choose 1} = (k+1)^2$$

And, that's as far as I got.

I get stuck with the ${k+1 \choose 2}$ part.

Answer 1

It's much simpler to give a direct proof: $$ 2 \binom{n}{2} + \binom{n}{1} = 2 \cdot \frac{n(n-1)}{2} + n = n^{2} - n + n = n^{2}. $$

Answer 2

We can also give this a combinatorial interpretation:

• $\binom{n}{2}$ is the number of unordered pairs of distinct elements of $\underline{n}$
• So $2\binom{n}{2}$ is the number of ordered pairs of distinct elements of $\underline{n}$

• So $2\binom{n}{2}+\binom{n}{1}$ is the number of ordered pairs of $\underline{n}$

• So $2\binom{n}{2}+\binom{n}{1}=|\underline{n}^2|$

• So $2\binom{n}{2}+\binom{n}{1}=n^2$

With enough care, this can be made into a rigorous proof (these are called "combinatorial proofs").