Question

50的三次方+51的三次方+......+99的三次方+100的三次方？

Answer 1

简便计算50^3+51^3+…+99^3+100^3
=1^3+2^3+…+99^3+100^3-(1^3+2^3+…+48^3+49^3)
=(1+2+…+99+100)^2-(1+2+…+48+49)^2
=[(1+100)x100/2]^2-[(1+49)x49/2]^2
=5050^2-1225^2
=25502500-1500625

=24001875

Answer 2

50³+51³+....+100³

=(1+2+3+……+100)²-(1+2+3+……+49)²

=[(1+100)×100÷2]²-[(1+49)×49÷2]²

=5050²-1225²

=25502500-1500625

=24001875
希望能帮到你

Answer 3

解:n1=49，n2=100，
S=n2^2(n2+1)/4一n1^2(n1+1)^2/4，
=100^2x101^2/4-49^2x50^2/4，
=25502500一6002500，
=19500000

追答

补充n2^2(n2+1)^2/4吊了应补上，特此说明。

计算错误特此更正:
(n2)^2(n2+1)^2/4-(n1)^2(n1+1)^2/4，
=100^2ⅹ101^2/4-49^2ⅹ50^2/4，
=25502500-1500625=24001875

Answer 4

50³+51³+52³+53³+…+99³+100³
=1³⁺2³+3³+…100³-(1³+2³+3³³+…+49³)
=1/4×100²×(100+1)²⁻1/4×49²×(49+1)²
=1/4×10000×10201-1/4×2401×2500
=25502500-1500625
=24001875

Answer 5

1³＋2³＋……＋n³=[n(n＋1)]²/4
所以所求=(100×101)²/4–(49×50)²/4
=24001875