∀n∈Nn∑k=1k(k+1)!=1−1(n+1)!
n=1 icin esitlik dogru.n≥1 icin dogru ise n+1 icin de dogru: n+1∑k=0k(k+1)!=n∑k=0k(k+1)!+n+1(n+2)! =1−1(n+1)!+n+1(n+2)!=1−1(n+2)!.
Tümevarım ne demek?