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).

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

it's not true: $$\binom20+2\ne\binom30$$ – J. W. Tanner Dec 03 '19 at 06:25 — J. W. Tanner, Dec 03 '19 at 06:25
What you have tried? – Kavi Rama Murthy Dec 03 '19 at 06:26 — Kavi Rama Murthy, Dec 03 '19 at 06:26

score 3 · Answer 1 · answered Dec 03 '19 at 06:26

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.

score 3 · Accepted Answer · answered Dec 03 '19 at 06:27

3

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

is just

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

answered Dec 03 '19 at 06:27

J. W. Tanner

60,406

thanks! i kept confusing myself with the $(n-2)!$ divisor, forgetting that it is effectively just truncating the factorial in the numerator. – tau Dec 03 '19 at 06:31

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

2 Answers2