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1.1^3+2^3+…+9^3+10^3
1^3+2^3+…+9^3+10^3
=Σ(k=1 to 10)_k^3
=10^2*(10+1)^2/4
=3025

2.1^3+2^3+…+99^3+100^3
1^3+2^3+…+99^3+100^3
=Σ(k=1 to 100)_k^3
=100^2*(100+1)^2/4
=25502500

3.1^3+2^3+…+999^3+1000^3
Sol
1^3+2^3+…+999^3+1000^3
=Σ(k=1 to 1000)_k^3
=1000^2*(1000+1)^2/4
=250500250000

您好，我是 lop，高興能解答您的問題。

可用公式 1^3 + 2^3 + 3^3 + ... + (n-1)^3 + n^3 = [ n(n+1)/2 ]^2。

(1)

1^3 + 2^3 + 3^3 + ... + 9^3 + 10^3

從 1^3 + 2^3 + 3^3 + ... + (n-1)^3 + n^3 = [ n(n+1)/2 ]^2 得

1^3 + 2^3 + 3^3 + ... + 9^3 + 10^3
= [10(10+1)/2]^2 ( 此時 n = 10 )
= [(10)(11)/2]^2
= 55^2
= 3025

(2)

1^3 + 2^3 + 3^3 + ... + 99^3 + 100^3

從 1^3 + 2^3 + 3^3 + ... + (n-1)^3 + n^3 = [ n(n+1)/2 ]^2 得

1^3 + 2^3 + 3^3 + ... + 99^3 + 100^3
= [100(100+1)/2]^2 ( 此時 n = 100 )
= [(100)(101)/2]^2
= 5050^2
= 25502500

(3)

1^3 + 2^3 + 3^3 + ... + 999^3 + 1000^3

從 1^3 + 2^3 + 3^3 + ... + (n-1)^3 + n^3 = [ n(n+1)/2 ]^2 得

1^3 + 2^3 + 3^3 + ... + 999^3 + 1000^3
= [1000(1000+1)/2]^2 ( 此時 n = 1000 )
= [(1000)(1001)/2]^2
= 500500^2
= 250500250000

If you want to find the sum , you may use the formula :
1^3 + 2^3 + ... + n^3 = [ n(n+1)/2 ]^2