3×(1×2+2×3+3×4+4×5...98×99+99×100)等于
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3×(1×2+2×3+3×4+4×5...98×99+99×100)等于
3×(1×2+2×3+3×4+4×5...98×99+99×100)等于
3×(1×2+2×3+3×4+4×5...98×99+99×100)等于
令an=n(n+1)=n^2+n,
Sn=1^2+2^2+……+n^2+1+2+……+n=(1/6)n(n+1)(2n+1)+n(n+1)/2
所以3×(1×2+2×3+3×4+4×5...98×99+99×100)
=3×【(1/6)*99*100*199+99*100/2】
=3×333300
=999900
1*2+2*3+3*4+4*5+……+98*99+99*100
= 1*(1+1)+2*(2+1)+3(3+1)+4(4+1)+……+98(98+1)+99(99+1)
=(1^2+1)+(2^2+2)+(3^2+3)+(4^2+4)+……+(98^2+98)+(99^2+99)
=(1^2+2^2+……+99^2)+(1+2+3+……+99)
=[99(99+1)(2*99+1)]/6 +[99(99+1)/2]
=333300
再乘以3就ok了
=1*(1+1)+2*(2+1)+3(3+1)+4(4+1)+……+98(98+1)+99(99+1)
=(1^2+1)+(2^2+2)+(3^2+3)+(4^2+4)+……+(98^2+98)+(99^2+99)
=(1^2+2^2+……+99^2)+(1+2+3+……+99)
=[99(99+1)(2*99+1)]/6 +[99(99+1)/2]
=333300