Sei \( \sum \limits_{k=1}^{n} k \cdot k !=(n+1) !-1 \)
Dann musst du zeigen
\( \sum \limits_{k=1}^{n+1} k \cdot k !=(n+2) !-1 \) .
Also los: \( \sum \limits_{k=1}^{n+1} k \cdot k ! \)
\( \sum \limits_{k=1}^{n} k \cdot k ! + (n+1)*(n+1)! \)
= (n+1)! - 1 + (n+1)*(n+1)!
= (n+1)! - 1 + n*(n+1)! + 1*(n+1)!
= 2*(n+1)! + n*(n+1)! - 1
= (2+n)*(n+1)! - 1
= (n+2)! - 1 q.e.d.