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10 Answers
62S=\sum_{r=0}^{n}\left\{{\sum_{s=r+1}^{n}{(r+s)(Cr+Cs+CrCs)}} \right\} \\ =\sum_{r=0}^{n}\left\{{\sum_{s=r}^{n}{(r+s)(Cr+Cs+CrCs)}} \right\}-\sum_{r=0}^{n}\left\{{\sum_{s=r}^{r}{(r+s)(Cr+Cs+CrCs)}} \right\} \\ =\sum_{r=0}^{n}\left\{{\sum_{s=r}^{n}{(r+s)(Cr+Cs+CrCs)}} \right\}-2\sum_{r=0}^{n}\left\{{r(2Cr+{C_r}^2)} \right\}
2S=\sum_{r=0}^{n}\left\{{\sum_{s=0}^{n}{(r+s)(Cr+Cs+CrCs)}} \right\}-2\sum_{r=0}^{n}\left\{{r(2Cr+{C_r}^2)} \right\}
2S=\sum_{r=0}^{n}\left\{{\sum_{s=0}^{n}{(r+n-s)(Cr+Cs+CrCs)}} \right\}-2\sum_{r=0}^{n}\left\{{r(2Cr+{C_r}^2)} \right\}
summing the above 2 ....
4S=\sum_{r=0}^{n}\left\{{\sum_{s=0}^{n}{(2r+n)(Cr+Cs+CrCs)}} \right\}-4\sum_{r=0}^{n}\left\{{r(2Cr+{C_r}^2)} \right\}
4S=\sum_{r=0}^{n}\left\{(2r+n){\sum_{s=0}^{n}{(Cr+Cs+CrCs)}} \right\}-4\sum_{r=0}^{n}\left\{{r(2Cr+{C_r}^2)} \right\}
4S=\sum_{r=0}^{n}\left\{(2r+n){{(n\times C_r+2^n+2^nC_r)}} \right\}-4\sum_{r=0}^{n}\left\{{r(2C_r+{C_r}^2)} \right\}
This is still a few steps far.. but see if i have done the right calculations and if you can manage it..
1Hint :
2\sum_{0\leq r<s\leq n}^{}\sum{}{CrCs} = \sum_{0\leq r,s\leq n}^{}\sum{}{Cr.Cs} - \sum_{0\leq r=s\leq n}^{}\sum{}{CrCs}
Solve three terms individually using this condition and u'll get answers for each in not more than 5-steps !
The only hindrance while solving this comes for the expression
\sum_{0\leq r=s\leq n}^{}{}\sum{(r+s)CrCs} = 2\sum_{r}^{}{r (Cr)^{2}}
which needs experts touch. nishant bhaiya, kaymant sir, prophet sir please help in reducing this expression !!!
1thank u nishant sir your solution is correct.
Do u have an easier solution?????