k*k! = (k+1-1)*k! = (k+1)*k! - k! = (k+1)! - k!
Scriem suma astfel (este o suma telescopica) :
2! - 1!
3! - 2!
4! - 3!
...
(n+1)! - n!
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Se reduc termenii pe diagonala ⇒ S = (n+1)! - 1
Salut,
[tex]\displaystyle S=\sum\linits_{k=1}^nk\cdot k!=\sum\linits_{k=1}^n (k+1-1)\cdot k!=\\\\=\sum\linits_{k=1}^n(k+1)\cdot k!-k!=\sum\linits_{k=1}^n(k+1)!-k!=\\\\=2!-1!+\\\\+3!-2!+\\\\+\ldots+\\\\+n!-(n-1)!+\\\\+(n+1)!-n!=(n+1)!-1!=(n+1)!-1,\ deci\ \boxed{\ S=(n+1)!-1.\ }[/tex]
Ai înțeles ?
Green eyes.
