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Knowing that $$sum_i=1^n i = fracn(n+1)2$$
how can I get closed form formula for
$$sum_i=1^n sum_j=1^i j$$
or
$$sum_i=1^n sum_j=1^i sum_k=1^j k$$
or any x times neasted summation like above

Let $f_k(n)$ be the closed form of the summation nested $k$ times. We know that
$$f_1(n)=frac12n(n+1)=binomn+12$$
$$f_k(n)=sum_j=1^n f_k-1(j)$$
So for the next function $f_2(n)$ we have
$$f_2(n)=sum_j=1^nbinomj+12=sum_j=2^n+1binomj2=binomn+23$$
By using the Hockey-stick identity (credits to Jean-Claude Arbaut).
Similarly for the next function $f_3(n)$ we have
$$f_3(n)=sum_j=1^nbinomj+23=sum_j=3^n+2binomj3=binomn+34$$
So one could conjecture that
$$f_k(n)=binomn+kk+1$$
which can be easily proven by induction as follows
$$f_k(n)=sum_j=1^nbinomj+k-1k=sum_j=k^n+k-1binomjk=binomn+kk+1$$
Hence we have that
$$boxedf_k(n)=binomn+kk+1=frac1(k+1)!n(n+1)(n+2)dots(n+k-1)(n+k)$$

We can write the last multiple sum as
beginalign*
colorbluesum_i_1=1^nsum_i_2=1^i_1sum_i_3=1^i_2i_3
&=sum_i_1=1^nsum_i_2=1^i_1sum_i_3=1^i_2sum_i_4=1^i_3 1\
&=sum_1leq i_4leq i_3leq i_2leq i_1leq n1tag1\
&,,colorblue=binomn+34tag2
endalign*
In (1) we observe the index range is the number of ordered $4$-tuples with repetition from a set with $n$ elements resulting in (2).

Here's a combinatorial way of thinking about it: first of all, note that we can go one level deeper and represent the innermost piece ($j$, or $k$, etc.) in your formulae as $sum_h=1^j1$; this means that the formula start to look like $displaystylesum_m=1^n1 =n$, $displaystylesum_m=1^nsum_l=1^m1=n(n+1)/2=n+1choose 2$, $displaystylesum_m=1^nsum_l=1^msum_k=1^l1=n+2choose 3$, etc. Now, let's look at what the left hand side is counting. In the first case, we're just counting the number of ways to choose an $m$ between $1$ and $n$ (inclusive); this is, self-evidently, just $n$. In the second, we're choosing a number $m$ between $1$ and $n$ inclusive, again, but then choosing an $l$ between $1$ and $m$; this is exactly the number of ways of choosing two numbers between $1$ and $n$, where we don't care about the order — that is, choosing $2$ and $5$ is exactly the same as choosing $5$ and $2$. Similarly, $displaystylesum_m=1^nsum_l=1^msum_k=1^l1$ counts the number of ways of choosing three numbers between $1$ and $n$, without regard to order; this is because we can sort the numbers we've chosen (since we don't care about order), and then note that the largest can be anywhere between $1$ and $n$, but then the next largest can only be between $1$ and the largest, etc.

Now, the difference between this and regular combinations is that in a regular combination every chosen number must be distinct; but if we have an ordered list $langle k, l, mrangle$ of the (not necessarily distinct) numbers we've chosen between $1$ and $n$ then we can turn this into an ordered list of not necessarily distinct numbers between $1$ and $n+2$: let $k'=k$, $l'=l+1$, $m'=m+2$. You should be able to convince yourself that this is a one-to-one correspondence between not-necessarily-distinct choices in $1ldots n$ and distinct choices in $1ldots n+2$, and the same principle extends to any number of choices. (This wikipedia link has more details).

$$S_n_2=sum_i=1^nsum_j=1^ij=sum_i=1^nfraci(i+1)2=frac12sum_i=1^ni^2+i=frac12left[fracn(n+1)(2n+1)6+fracn(n+1)2right]=fracn(n+1)(n+2)6$$
and now:
$$S_n_3=sum_i=1^nsum_j=1^isum_k=1^jk=frac16sum_i=1^ni(i+1)(i+2)=frac16sum_i=1^ni^3+3i^2+2i=frac16left[fracn^2(n+1)^24+fracn(n+1)(2n+1)2+n(n+1)right]=fracn(n+1)6left[fracn(n+1)4+frac(2n+1)2+1right]=fracn(n+1)(n+2)(n+3)24$$
and we can see a pattern here. For a series $S_n_a$ with $a$ nested summations the following is true:
$$S_n_a=frac1(a+1)!prod_b=0^a(n+b)=frac(n+a)!(n-1)!(a+1)!$$