Prove that using induction that $\binom22+\dots+\binom n2 = \binom{n+1}2$

Question

so I have this math problem where I have to prove this using induction. $$\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}n\\2\end{pmatrix}=\begin{pmatrix}n+1\\3\end{pmatrix}\text{ for } n\ge2$$

I start by checking the initial case $P(2)=\begin{pmatrix}2\\2\end{pmatrix}=1$ It's good.

I then assume P(k) is true. $P(k)=\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}k\\2\end{pmatrix}=\begin{pmatrix}k+1\\3\end{pmatrix}$

I then try to prove P(k+1) is also true. $P(k+1)=\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}k+1\\2\end{pmatrix}=\begin{pmatrix}k+2\\3\end{pmatrix}$

I know that $\begin{pmatrix}k+1\\3\end{pmatrix}=\begin{pmatrix}k\\2\end{pmatrix}+\begin{pmatrix}k\\3\end{pmatrix}$

Now I'm stuck here... I'm not sure how to finish this proof. Thanks.

Answer 1

To prove $P(k+1)$, start with your left hand side, i.\,e. $$ \def\P#1{\binom 22 + \binom 32 + \cdots + \binom{#1}2}\P{k+1}$$ Write it as $$ \P k + \binom{k+1}2 \tag+ $$ Now use, that by the induction hypothesis, we have $$ P(k): \P k = \binom{k+1}3 $$ Hence, we can write $(+)$ as $$ \P k + \binom{k+1}2 = \binom{k+1}3 + \binom{k+1}2 $$ But the right hand side here equals (as you write - but for $k+1$ instead of $k$) $$ \binom{k+1}3 + \binom{k+1}2 = \binom{k+2}3 $$ Collecting everything, you are done.

Answer 2

You said, $${k+1\choose3} = {k\choose2}+{k\choose3}$$

$$\implies {k+2\choose3} = {k+1\choose2}+{k+1\choose3}$$

Now use you induction hypothesis, i.e. $${2\choose2}+{3\choose2}+\dots+{k\choose2} = {k+1\choose3}$$

Substituting the value of ${k+1\choose3}$ in the above equation you get the desired result.