These were simple induction proofs, so I decided to try and prove them combinatorially. I think I nailed the first one, not so sure about the second one.
$\sum_{i=1}^n(i)(i!)=(n+1)!-1$
Have $n+1$ integers want all orderings except for strictly ascending. Can pick all permutations and subtract one(right hand side) or remove each # $\ge$ minimum # one at a time and place to the left of the ordering, gauranteeing they will never be in strictly ascending order(left hand side). There would be $n$ of these #'s times $n!$ for the ordering of the remaining #'s. $(i)(i!)$ 'occurs' $n$ times making it $n(n!)$.
$\sum_{i=1}^ni(2^i)=2+(n-1)2^{n+1}$
Have 2 groups of $n$ bits each. want # combinations of which bit(they are numbered/specified) selected from first group with all configurations of bits in $2nd$ group(they are not numbered/specified). LHS $i(2^i)$ 'occurs' $n$ times making it $n(2^n)$. This is how the bits are first placed. For RHS can remove one bit from first group and place it in the $2nd$ so that new total is the configuration$(1/0)$ of bit selected$(2)$ plus all combos of remaining bits in first group$(n-1)$ with all configs of bits in second group(now $2^{n+1}$).
Please verify proofs, especially second one. Thanks.
In the first one the lefthand side isn’t $$\underbrace{n!+n!+\ldots+n!}_{n\text{ times}}\;,$$ which is what you’ve described, but rather $$1\cdot1!+2\cdot2!+\ldots+n\cdot n!\;.$$ I can’t altogether follow your argument for the second one, but I think that you’ve made a similar mistake in interpreting the summation.
I’ve not yet come up with a combinatorial argument for the first identity, and I don’t really expect to, but I have one for the second. You have $n+1$ slips of paper, numbered $0$ through $n$, lined up on the table, white on one side and red on the other. All of them are white side up. Now choose a number $k\in\{1,\ldots,n\}$, and turn over slip $k$. Then turn over any subset of the $k$ slips to the left of slip $k$ (i.e., those with numbers less than $k$); you may choose to turn over none at all. Finally, choose any one of the $k$ slips to the left of slip $k$ and put a coin on it; it doesn’t matter whether the slip is white side up or red side up. For a fixed choice of $k$ there are $k2^k$ ways to do this, so there are $\sum_{k=1}^nk2^k$ ways altogether, and they all produce different outcomes.
Now imagine that instead you turn over any subset of the $n+1$ slips and put a coin on any slip except slip $n$. If there is no red slip to the right of the slip with the coin, turn over slip $n$. There are altogether $n2^{n+1}$ ways to do this, and the possible outcomes are exactly the same as for the first procedure. However, the outcome in which slip $n$ is the only red slip to the right of the coin is produced in two ways, one in which it was part of the set of slips originally chosen to be turned over, and one in which it was not. I claim that there are $2(2^n-1)$ such outcomes. If this is correct, the number of distinct outcomes is $n2^{n+1}-2(2^n-1)=(n-1)2^{n+1}+2$, and the identity is established.
Consider an outcome in which slip $n+1$ is the only red slip to the right of the coin. Let $S$ be the set of slips that either are red or have the coin; $S$ can be any non-empty subset of the first $n$ slips, so there are $2^n-1$ possibilities for $S$. Let $k$ be the number of the slip with the coin. Then $k=\max S$, and there are two possibilities: either slip $k$ is red, or it isn’t. Thus, there are indeed $2(2^n-1)$ outcomes of this type, and we’re done.