sum equals to...? - II

Alen0905

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Mar 22, 2020
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11
new one...seems a bit more complicated...
 

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MarkFL

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Nov 24, 2012
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Let's begin by writing:

\(\displaystyle S=\sum_{k=2}^{n}\left(k(k-1){n \choose k}\right)=\sum_{k=2}^{n}\left(k(k-1)\frac{n!}{k!(n-k)!}\right)=\sum_{k=2}^{n}\left(\frac{n!}{(k-2)!(n-k)!}\right)\)

Re-index:

\(\displaystyle S=\sum_{k=0}^{n-2}\left(\frac{n!}{k!(n-(k+2))!}\right)=\sum_{k=0}^{n-2}\left(\frac{n(n-1)(n-2)!}{k!((n-2)-k)!}\right)\)

Can you proceed?
 

Alen0905

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Joined
Mar 22, 2020
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11
Thanks! Yes, i think i can see it from there :)
 

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