Question

\[\sum \sum (r+s)\left(\binom{n}{r} + \binom{n}{s}\right)\]where\[0 \le r < s \le n\]

Answer 1

\[\sum_{r=0}^{n}\sum_{s=0}^n f (r,s)=\sum \sum f(r,r) + 2\sum_{0 \le r} \sum\limits_{< s \le n} f(r,s) \]

Answer 2

\[\sum_{s=0}^n \sum_{r=0}^n (r+s)(C_r + C_s) \]\[= \sum_{s=0}^{n}\left(\sum_{r=0}^n rC_r + s\sum_{r=0}^n C_r + C_s\sum_{r=0}^{n} r + sC_s \sum_{r=0}^n 1\right)\]\[= \sum_{s=0}^n\left( n\cdot 2^{n-1} + s\cdot 2^n +\frac{n(n+1)}{2}C_s + sC_s\right) \]

Answer 3

\[= n(n+1) 2^{n-1} + \frac{n(n+1)}{2} 2^n + \frac{n(n+1)}{2}2^n + n 2^{n-1}\]\[= 3n(n+1)2^n +n2^{n-1}\]\[= \cdots\]

Answer 4

I've done all this by memory so I hope I've got all the results right so far.

Answer 5

\[\sum_{r=0}^{n}(2r)(C_r + C_r) \]\[= 4\sum_{r=0}^n rC_r\]\[= n \cdot 2^{n+1}\]

Answer 6

Now we've got our sum as\[\frac{1}{2}\left(3n(n+1) 2^{n} + n2^{n-1} - n2^{n+1}\right)\]\[= 3n(n+1) 2^{n-1} + n 2^{n-2} - n2^{n}\]\[= (6n^2 + 6n + n - 4n)\cdot 2^{n-2}\]\[= (6n^2 + 3n)\cdot 2^{n-2}\]