Reducing this binomial expression

Question

I need help for showing that: $$\sum\limits_{k=2}^{50} = k \cdot(k-1)\binom{50}{k}$$

is equal to:

$$50\cdot 49\cdot 2^{48}$$ please help , thank you.

Answer 1

Hint. One may observe that $$ k\times(k-1)\times\binom{50}{k}=50\times49\times\binom{48}{k-2} $$ giving $$ \sum\limits_{k=2}^{50} k\times(k-1)\times\binom{50}{k}=50\times49\times\sum\limits_{k=2}^{50}\binom{48}{k-2}=50\times49\times\sum\limits_{m=0}^{48}\binom{48}{m} $$ then one may use the binomial theorem.

Answer 2

Here is a combinatorial proof:

A committee of $2 \le k \le 50$ people needs to be formed from $50$ people such that: There is exactly one president, exactly one vice president, and other members are optional. How many ways of forming such a committee?

Method 1:
Choosing president: $50$ ways.
Choosing vice-president $49$ ways.
All other members are optional, so $2^{48}$ ways of choosing them.

Method 2
On the other hand, fix the size of committee to be $2 \le k \le 50$.
Choose $k$ people from $50$: $\binom{50}{k}$ ways.
Now, out of $k$ people, choose $2$ people for president and vice president: $\binom{k}{2}$ ways.
Choose president and vice president from the $2$ people: 2 ways.
So total ways for fixed $k$: $\binom{50}{k}k (k-1)$.
Now sum this over all possible values of $k$ to get the quantity in the LHS.

Answer 3

Alternatively, consider the binomial expansion of $(1+x)^{50}$, and differentiate this twice, then substitute the value of $x=1$. The result follows immediately.