3N5+5N3+7N琌15计
N琌礛计玥3N5+5N3+7N琌15计
硈尿礛计縩(N-1)N(N+1)琌3计....(1)
安砞N=3k玥(N-1)N(N+1)=(N2-1)N=(9k2-1)(3k)=琌3计
安砞N=3k+1玥(N-1)N(N+1)=(N2-1)N=(9k2+6k)(3k+1)=3(3k2+2k)(3k+1)琌3计
安砞N=3k+2玥(N-1)N(N+1)=(N2-1)N=(9k2+12k+3)(3k+2)=3(3k2+4k+1)(3k+2)琌3计
┮硈尿礛计縩(N-1)N(N+1)琌3计
N琌礛计玥N(N2+1)(N+1)(N-1)琌5计....(2)
安砞N=5k玥N(N2+1)(N+1)(N-1)=(5k)(N2+1)(N+1)(N-1)琌5计
安砞N=5k+1玥N(N2+1)(N+1)(N-1)=N(N2+1)(N+1)(5k)琌5计
安砞N=5k+2玥N(N2+1)(N+1)(N-1)=N[(5k+2)2+1](N+1)(N-1)=N(25k2+20k+5)(N+1)(N-1)琌5计
安砞N=5k-1玥N(N2+1)(N+1)(N-1)=N(N2+1)(5k)(N-1)琌5计
安砞N=5k-2玥N(N2+1)(N+1)(N-1)=N[(5k-2)2+1](N+1)(N-1)=N(25k2-20k+5)(N+1)(N-1)琌5计
┮N琌礛计玥N(N2+1)(N+1)(N-1)琌5计
パ瓃(1)(2)
3N5+5N3+7N=3(N5-N)+5(N3-N)+15N=3N(N4-1)+5N(N2-1)+15N=3N(N2+1)(N+1)(N-1)+5N(N+1)(N-1)+15N琌15计
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