new one...seems a bit more complicated...
A Alen0905 New member Joined Mar 22, 2020 Messages 21 Mar 23, 2020 #1 new one...seems a bit more complicated... Attachments IMG_20200323_083735.jpg 2.1 MB · Views: 3
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Mar 23, 2020 #2 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?
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?
A Alen0905 New member Joined Mar 22, 2020 Messages 21 Mar 23, 2020 #3 Thanks! Yes, i think i can see it from there Attachments IMG_20200323_152655.jpg 3.6 MB · Views: 3