1
$\begingroup$

finding $\displaystyle \mathop{\sum\sum}_{0 \leq i < j \leq n}(i+j)\binom{n}{i}\binom{n}{j}$

expanding sum $\displaystyle (0+1)\binom{n}{0}\binom{n}{1}+(0+2)\binom{n}{0}\binom{n}{2}+\cdots \cdots +(0+n)\binom{n}{0}\binom{n}{n}+(1+2)\binom{n}{1}\binom{n}{2}+(1+3)\binom{n}{1}\binom{n}{3}+\cdots+(1+n)\binom{n}{1}\binom{n}{n}+\cdots +(n-1+n)\binom{n}{n-1}\binom{n}{n}$

wan,t be able to go further , some help me

  • 0
    I edited the question. Is this what you meant? – 2017-02-02
  • 0
    Related: http://math.stackexchange.com/questions/1705589/the-value-of-mathop-sum-sum-0-leq-i-j-leq-n-1i-j1-binomni-binom?rq=1. – 2017-02-02

1 Answers 1

6

You may use a diagonal argument. Observe that $$\sum_{0\leq ihere.

Therefore $$\sum_{0\leq i