Binomial coefficient identity ${n \choose 2} + n = {n + 1 \choose 2}$

Question

I can see why this is true using Pascal's triangle or the recurrence relation, but algebraically there must be a way and I'm just missing something (trying to sort out factorials using the binomial coefficient formula didn't work out for me).

${n \choose 2} + n = {n + 1 \choose 2}$

Answer 1

$${n \choose 2} + n = {n + 1 \choose 2}$$

is just

$$\dfrac{n(n-1)}2+n=\dfrac{n(n+1)}2$$

Answer 2

The correct identity is $$ \binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}. $$ To prove it, observe that $$ \binom{n}{k}=\frac{n-k}{n+1}\binom{n+1}{k+1}\qquad \text{and}\qquad \binom{n}{k+1}=\frac{k+1}{n+1}\binom{n+1}{k+1}. $$ Add together these two equations to get the result.