Combinatorial Proof of $n(n+1)2^{n-2} = \sum_{i=1}^ni^2\binom{n}{i}$

Question

$n(n+1)2^{n-2} = \sum_{i=1}^ni^2\binom{n}{i}$

I would like to prove above identity combinatorially.

$n(n+1)$ of LHS is doubled amount of summing up from $1$ to $n$ and remaining $2^{n-1}$ denotes that there's some $n -1$ consecutive choice of something being exist or not.

How could I relate this two factors into potential those of RHS ?

Or will there any other option that I can refer to?

Any advice would be appreciate.

Answer 1

Hint:

The right-hand side counts the number of ways to choose a subcommittee (of any non-empty size) from a committee of $n$ people, choose a leader, and choose a note-taker... with the understanding that one person can be both the leader and the note-taker.

Can you see how to show that the left-hand side counts the same? It might help to note that $n+1=(n-1)+2$.

Answer 2

Following Nick's hint I'd like to construct the RHS.

First ask each one of n people whether he/she wants to participate in sub-committee $ = 2^n$

Then I choose $2$ people among the entire $n$ to designate each one of $2$ into the leader or note-taker position = $\binom{n}{2}\cdot2$

but the possibility is only $1\over2^2$ which I successfully choose 2 people who both of them said "YES! I want to be a member of sub-committee with extra work!" =$\binom{n}{2}\cdot2\cdot{1\over4}$

Now, I want to consider $2$ positions only filled with one person. It's simple. Just take $1$ among $n$ with $1\over2$ of possibility = $n\over2$

Now, summing up, the total counts will be $2^n({n(n-1)\over4}+{n\over2})$ which is equal to LHS