C ++程序的前n个自然数的平方和

给定正整数N。任务是找到1 ² + 2 ² + 3 ² +….. + N ² 。

例子：

Input : N = 4
Output : 30
12 + 22 + 32 + 42
= 1 + 4 + 9 + 16
= 30

Iput : N = 5
Output : 55

方法1：O(N)的想法是运行一个从1到n的循环，对于每个i，1 <= i <= n，求和^2。

// CPP Program to find sum of square of first n natural numbers
#include 
using namespace std;
  
// Return the sum of the square 
// of first n natural numbers
int squaresum(int n)
{
    // Iterate i from 1 and n
    // finding square of i and add to sum.
    int sum = 0;
    for (int i = 1; i <= n; i++)
        sum += (i * i);
    return sum;
}
  
// Driven Program
int main()
{
    int n = 4;
    cout << squaresum(n) << endl;
    return 0;
}

输出：

30

方法2：O(1)

证明：

We know,
(k + 1)3 = k3 + 3 * k2 + 3 * k + 1
We can write the above identity for k from 1 to n:
23 = 13 + 3 * 12 + 3 * 1 + 1 ......... (1)
33 = 23 + 3 * 22 + 3 * 2 + 1 ......... (2)
43 = 33 + 3 * 32 + 3 * 3 + 1 ......... (3)
53 = 43 + 3 * 42 + 3 * 4 + 1 ......... (4)
...
n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ......... (n - 1)
(n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ......... (n)

Putting equation (n - 1) in equation n,
(n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1
         = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1

By putting all equation, we get
(n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1
n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2
n * (n + 1) * (2 * n + 1)/2  = 3 * Σ k2
n * (n + 1) * (2 * n + 1)/6  = Σ k2

// CPP Program to find sum
// of square of first n
// natural numbers
#include 
using namespace std;
  
// Return the sum of square of
// first n natural numbers
int squaresum(int n)
{
    return (n * (n + 1) * (2 * n + 1)) / 6;
}
  
// Driven Program
int main()
{
    int n = 4;
    cout << squaresum(n) << endl;
    return 0;
}

输出：

30

避免过早溢出：
对于大n，(n *(n + 1)*(2 * n + 1))的值将溢出。我们可以使用n *(n + 1)必须被2整除的事实在一定程度上避免溢出。

// CPP Program to find sum of square of first
// n natural numbers. This program avoids
// overflow upto some extent for large value
// of n.
#include 
using namespace std;
  
// Return the sum of square of first n natural
// numbers
int squaresum(int n)
{
    return (n * (n + 1) / 2) * (2 * n + 1) / 3;
}
  
// Driven Program
int main()
{
    int n = 4;
    cout << squaresum(n) << endl;
    return 0;
}

输出：

30

有关更多详细信息，请参阅关于第n个自然数的平方和的完整文章！

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