程序来查找级数之和(1 * 1)+(2 * 2)+(3 * 3)+(4 * 4)+(5 * 5)+…+(n * n)

您得到了一系列(1 * 1)+(2 * 2)+(3 * 3)+(4 * 4)+(5 * 5)+…+(n * n)，求出系列直到第n个学期。
例子 ：

Input : n = 3
Output : 14
Explanation : 1 + 1/2^2 + 1/3^3

Input : n = 5
Output : 55
Explanation : (1*1) + (2*2) + (3*3) + (4*4) + (5*5)

C++

// CPP program to calculate the following series
#include
using namespace std;
 
// Function to calculate the following series
int Series(int n)
{
    int i;
    int sums = 0;
    for (i = 1; i <= n; i++)
        sums += (i * i);
    return sums;
}
 
// Driver Code
int main()
{
    int n = 3;
    int res = Series(n);
    cout<

---

C

// C program to calculate the following series
#include 
 
// Function to calculate the following series
int Series(int n)
{
    int i;
    int sums = 0;
    for (i = 1; i <= n; i++)
        sums += (i * i);
    return sums;
}
 
// Driver Code
int main()
{
    int n = 3;
    int res = Series(n);
    printf("%d", res);
}

Java

// Java program to calculate the following series
import java.io.*;
class GFG {
 
    // Function to calculate the following series
    static int Series(int n)
    {
        int i;
        int sums = 0;
        for (i = 1; i <= n; i++)
            sums += (i * i);
        return sums;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 3;
        int res = Series(n);
        System.out.println(res);
    }
}

Python

# Python program to calculate the following series
def Series(n):
    sums = 0
    for i in range(1, n + 1):
        sums += (i * i);
    return sums
 
# Driver Code
n = 3
res = Series(n)
print(res)

C#

// C# program to calculate the following series
using System;
class GFG {
 
    // Function to calculate the following series
    static int Series(int n)
    {
        int i;
        int sums = 0;
        for (i = 1; i <= n; i++)
            sums += (i * i);
        return sums;
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 3;
        int res = Series(n);
        Console.Write(res);
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// Javascript program to calculate the following series
 
    // Function to calculate the following series
    function Series( n) {
        let i;
        let sums = 0;
        for (i = 1; i <= n; i++)
            sums += (i * i);
        return sums;
    }
 
    // Driver Code
      
        let n = 3;
        let res = Series(n);
        document.write(res);
 
// This code contributed by Princi Singh
 

输出 ：

14

时间复杂度：O(n)
请参考下面的文章O(1)解决方案。
前n个自然数的平方和