[Project Euler] 来做欧拉项目练习题吧: 题目006
2011-01-22 08:55
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[Project Euler] 来做欧拉项目练习题吧: 题目006
周银辉
问题描述:
The sum of the squares of the first ten natural numbers is,
12 + 22 + ... + 102 = 385The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)2 = 552 = 3025Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025
385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
问题分析:
嗯,这是个纯数学题了,有数学公式的:
(1+2+3...+n)^2 = ((1+n)*n/2)^2
(1^2 + 2^2 + 3^2 +...+ n^2) = 1/6 * n(n+1)(2n+1)
1+2+3+...+n这是等差数列,所以(1+n)*n/2
而(1^2 + 2^2 + 3^2 +...+ n^2) 嘛,可以这样推导:
首先,
n*(n+1)*(2n+1) - (n-1)*n*(2n-1)= (n^2+n)*(2n+1) - (n^2-n)*(2n-1)
= (2*n^3+n^2+2*n^2+n) - (2*n^3-n^2-2*n^2+n)
= 2*n^3+3n^2+n - 2*n^3+3*n^2-n
= 6*n^2
然后,
当 n=1时, 1*2*3 - 0 = 6*1^2
当 n=2时, 2*3*5 - 1*2*3 = 6*2^2
当 n=3时, 3*4*7 - 2*3*5 = 6*3^2
...
当 n=n-1时,(n-1)*n*(2n-1) - (n-2)*(n-1)*(2n-3) = 6*(n-1)^2
当 n=n时, n*(n+1)*(2n+1) - (n-1)*n*(2n-1) = 6*n^2
上面的各个等式左右对应相加,左边的很多项会正负抵消,然后得到:
n*(n+1)*(2n+1) = 6*(1^2+2^2+3^2+...+n^2)
所以,(1^2+2^2+3^2+...+n^2) = n(n+1)(2n+1)/6
注:当完成题目后,对于某些题,官方网站会给出参考答案,在我的博客里不会将官方答案贴出来,仅仅会写下我自己当时的思路,除非两者不谋而合。另外,如果你有更好的思路,请留言告诉我,我非常乐意参与到讨论中来。
周银辉
问题描述:
The sum of the squares of the first ten natural numbers is,
12 + 22 + ... + 102 = 385The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)2 = 552 = 3025Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025
385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
问题分析:
嗯,这是个纯数学题了,有数学公式的:
(1+2+3...+n)^2 = ((1+n)*n/2)^2
(1^2 + 2^2 + 3^2 +...+ n^2) = 1/6 * n(n+1)(2n+1)
1+2+3+...+n这是等差数列,所以(1+n)*n/2
而(1^2 + 2^2 + 3^2 +...+ n^2) 嘛,可以这样推导:
首先,
n*(n+1)*(2n+1) - (n-1)*n*(2n-1)= (n^2+n)*(2n+1) - (n^2-n)*(2n-1)
= (2*n^3+n^2+2*n^2+n) - (2*n^3-n^2-2*n^2+n)
= 2*n^3+3n^2+n - 2*n^3+3*n^2-n
= 6*n^2
然后,
当 n=1时, 1*2*3 - 0 = 6*1^2
当 n=2时, 2*3*5 - 1*2*3 = 6*2^2
当 n=3时, 3*4*7 - 2*3*5 = 6*3^2
...
当 n=n-1时,(n-1)*n*(2n-1) - (n-2)*(n-1)*(2n-3) = 6*(n-1)^2
当 n=n时, n*(n+1)*(2n+1) - (n-1)*n*(2n-1) = 6*n^2
上面的各个等式左右对应相加,左边的很多项会正负抵消,然后得到:
n*(n+1)*(2n+1) = 6*(1^2+2^2+3^2+...+n^2)
所以,(1^2+2^2+3^2+...+n^2) = n(n+1)(2n+1)/6
注:当完成题目后,对于某些题,官方网站会给出参考答案,在我的博客里不会将官方答案贴出来,仅仅会写下我自己当时的思路,除非两者不谋而合。另外,如果你有更好的思路,请留言告诉我,我非常乐意参与到讨论中来。
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