sum equals to...? - II
- Thread starter Alen0905
- Start date
- Joined
- Nov 24, 2012
- Messages
- 3,021
Let's begin by writing:
[MATH]S=\sum_{k=2}^{n}\left(k(k-1){n \choose k}\right)=\sum_{k=2}^{n}\left(k(k-1)\frac{n!}{k!(n-k)!}\right)=\sum_{k=2}^{n}\left(\frac{n!}{(k-2)!(n-k)!}\right)[/MATH]
Re-index:
[MATH]S=\sum_{k=0}^{n-2}\left(\frac{n!}{k!(n-(k+2))!}\right)=\sum_{k=0}^{n-2}\left(\frac{n(n-1)(n-2)!}{k!((n-2)-k)!}\right)[/MATH]
Can you proceed?
[MATH]S=\sum_{k=2}^{n}\left(k(k-1){n \choose k}\right)=\sum_{k=2}^{n}\left(k(k-1)\frac{n!}{k!(n-k)!}\right)=\sum_{k=2}^{n}\left(\frac{n!}{(k-2)!(n-k)!}\right)[/MATH]
Re-index:
[MATH]S=\sum_{k=0}^{n-2}\left(\frac{n!}{k!(n-(k+2))!}\right)=\sum_{k=0}^{n-2}\left(\frac{n(n-1)(n-2)!}{k!((n-2)-k)!}\right)[/MATH]
Can you proceed?
