<s>[INST] Proving $\binom{n}{m}+2\binom{n-1}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}$ 
For $m,n\in\mathbb{N},\;n\geq m$, prove the following:
$$
\tag{i}\binom{n}{m}+\binom{n-1}{m}+\binom{n-2}{m}+......+\binom{m}{m} = \binom{n+1}{m+1}
$$
  $$
\tag{ii}\binom{n}{m}+2\binom{n-1}{m}+3\binom{n-2}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}
$$

My Attempt:
For $(\mathrm{i})$, We can write $\binom{n}{m}$ as the coefficient of $x^m$ in $(1+x)^n$. Thus we can also write $\binom{n-1}{m}$ as the coefficient of $x^m$ in $(1+x)^{n-1}$ and $\binom{n-2}{m}$ as the coefficient of $x^m$ in $(1+x)^{n-2}$. So we have to find the coefficient of $x^m$ in
$$
(1+x)^n+(1+x)^{n-1}+(1+x)^{n-2}+........+(1+x)^{m}
$$
Using the formula for the sum of a geometric progression, this sum equals
$$\frac{(1+x)^{n+1}-(1+x)^m}{x}$$
So we now need to find the coefficient of $x^m$ in
$$
\frac{(1+x)^{n+1}-(1+x)^m}{x}
$$
or, equivalently, we need to find the coefficient of $x^{m+1}$ in
$$
(1+x)^{n+1}-(1+x)^m = \binom{n+1}{m+1}
$$
We can use a similar method to solve $(\mathrm{ii})$. Can these questions be solved using combinatorial methods instead?
 [/INST] Suppose $n \geq m$ and there is a set $S$ of $n + 2$ objects, denoted as $o_1, o_2, \cdots, o_{n+2}$. Your task to sample $m + 2$ objects from $S$. Let $X_i$ denote the # of ways to choose $m + 2$ objects such that the object with second minimum id is $o_i$. Easy to see that
$$
X_i = (i-1) \cdot { n + 2 - i \choose m }
$$
So we have
$$
\sum_{i=2}^{n+2-m} X_i = {n \choose m} + 2\cdot{n-1 \choose m} + \cdots + (n + 1 -m) \cdot {m \choose m} = {n + 2 \choose m + 2}
$$

Similar derivation can be applied to (i) with some minor modification. Instead of enumerating the second minimum id, you need to enumerate the minimum id instead.
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