1×2+2×3+3×4+...+2002*2003

1×2+2×3+3×4+...+2002*2003
=1x(1+1)+2x(2+1)+3x(3+1)+...+2002x(2002+1)
=(1^2+2^2+3^2+...+2002^2)+(1+2+3+...+2002)
=[2002x(2002+1)(2x2002+1)/6]+[(2002+1)x2002/2]
=2676679005+2005003
=2678684008

1×2+2×3+3×4+...+2002×2003
=1×(1+1)+2×(2+1)+3×(3+1)+...+2002×(2002+1)
=(1^2+2^2+3^2+...+2002^2)+(1+2+3+...+2002)
=[2002×(2002+1)(2×2002+1)/6]+[(2002+1)×2002/2]
=2676679005+2005003
=2678684008

:An=n(n+1);
求和公式为:Sn=1/3*n(n+1)(n+2);
本题以n=2002代入求和公式即得答案:
S=1/3*2002×2003×2004
=2678684008.

1×2+2×3+3×4+...+2002×2003=1×(1+1)+2×(2+1)+3×(3+1)+...+2002×(2002+1)=(1^2+2^2+3^2+...+2002^2)+(1+2+3+...+2002)=[2002×(2002+1)(2×2002+1)/6]+[(2002+1)×2002/2]=2676679005+2005003=2678684008.注:...