Finding $\sum_{_{0\le i}}\sum_{_{

Question

Let $$S=\sum_{0\le i}\sum_{<j\le n} {n \choose i}\tag{*}$$ Consider the summation: $$\sum_{j=0}^{n}\sum_{i=0}^{n} A_i B_j=\sum_{i=0}^n A_i B_i+\sum_{0\le i<j\le n} A_i B_j+ \sum_{n\ge i>j\ge 0} A_i B_j$$ If $B_j=1$, then the second and the third sums in RHS above are equal, then $$\sum_{0\le i<j\le n}A_i=\frac{1}{2}\left[\sum_{j=0}^n\sum_{i=0}^{n} A_i-\sum_{i=0}^n A_i\right]=\frac{1}{2}\left[(n+1)\sum_{i=0}^nA_i-\sum_{i=0}^nA_i\right]$$ Finally, taking $A_i={n \choose i}$ and $\sum_{i=0}^{n} {n \choose i} =2^n$, we get $S=n2^{n-1}$.

The question is: How else this sum (*) can be evaluated?

Edit: It requires good steps to show that $S=\sum_{k=0}^{n} k {n \choose k}$ and hence this question cannot be taken as duplicate. See the positive comments of @jjagmath and @Phicar below.

@Aig It is a restricted double summation. I have edited it now. — Z Ahmed, Mar 13 '24 at 15:25
Well, you can write your sum as $$\sum _{i = 0}^n\binom{n}{i}(n-i),$$ where the $n-i$ comes from all choises of $j$ given $i$. Just use that $\binom{n}{m}\binom{m}{k}=\binom{n}{k}\binom{n-k}{m-k}$. — Phicar, Mar 13 '24 at 15:33
Does this answer your question? How to compute this combinatoric sum? $$\sum_{i = 0}^n\binom{n}{i}(n-i)=\sum_{k= 0}^n\binom n k k$$ and there are many post about this sum. — Anne Bauval, Mar 13 '24 at 15:36
Is your question correct? Isnt it it supposed to be Double sigma 0<i<j<n nci ncj? I think your question is missing the ncj term — Maths lover, Mar 13 '24 at 15:42
The fun is that the present proposal strikes interest, some even find it wrong or misprinted! — Z Ahmed, Mar 13 '24 at 15:50

score 2 · Accepted Answer · answered Mar 13 '24 at 16:09

2

This sum is actually a double sum $$S=\sum_{j=0}^{n} \sum_{i=0}^{j-1} {n \choose i}$$ This can be expanded as $$S={n \choose 0}+\left[{n \choose 0}+{n \choose 1}\right]+\left[{n \choose 0}+{n\choose 1}+{n \choose 2}\right]+\left[{n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}\right]+.......+[0]$$ This is as though the frequency of ${n \choose i}$ is $(n-i)$, then $$=\sum_{i=0}^{n} (n-i) {n \choose i}=\sum_{k=0}^n k {n \choose k}=n 2^{n-1}$$

answered Mar 13 '24 at 16:09

MathDona

632

1

The main step (the previous ones were already noticed in comments) is the last equality, which (as also commented) is a mega duplicate. – Anne Bauval Mar 13 '24 at 16:12

Finding $\sum_{_{0\le i}}\sum_{_{

1 Answers1