2,10，30，68，…系列的前N个项的总和。

给定数字N，任务是找到以下系列的前N个项的总和：

Sn = 2 + 10 + 30 + 68 + … upto n terms

例子：

Input: N = 2
Output: 12
2 + 10
= 12

Input: N = 4 
Output: 40
2 + 10 + 30 + 68
= 110

方法：让第n个项用tn表示。
可以通过如下拆分每个术语来轻松解决此问题：

Sn = 2 + 10 + 30 + 68 + ......
Sn = (1+1^3) + (2+2^3) + (3+3^3) + (4+4^3) +......
Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^3 + 2^3 + 3^3 + 4^3 + ...upto n terms)

我们观察到Sn可以分解为两个系列的总和。
因此，前n个项的总和如下：

Sn = (1 + 2 + 3 + 4 + ...unto n terms) + (1^3 + 2^3 + 3^3 + 4^3 + ...upto n terms)
Sn = n*(n + 1)/2 + (n*(n + 1)/2)^2

下面是上述方法的实现：

C++

// C++ program to find sum of first n terms
#include 
using namespace std;
 
// Function to calculate the sum
int calculateSum(int n)
{
 
    return n * (n + 1) / 2
           + pow((n * (n + 1) / 2), 2);
}
 
// Driver code
int main()
{
    // number of terms to be
    // included in the sum
    int n = 3;
 
    // find the Sum
    cout << "Sum = " << calculateSum(n);
 
    return 0;
}

Java

// Java program to find sum of first n terms
 
public class GFG {
     
    // Function to calculate the sum
    static int calculateSum(int n)
    {
       
        return n * (n + 1) / 2 
               + (int)Math.pow((n * (n + 1) / 2), 2);
    }
     
    // Driver code
    public static void main(String args[])
    {
        // number of terms to be
        // included in the sum
        int n = 3;
       
        // find the Sum
        System.out.println("Sum = "+ calculateSum(n));
    }
    // This Code is contributed by ANKITRAI1
}

Python3

# Python program to find sum
# of first n terms
 
# Function to calculate the sum
def calculateSum(n):
    return (n * (n + 1) // 2 +
        pow((n * (n + 1) // 2), 2))
 
# Driver code
 
# number of terms to be
# included in the sum
n = 3
 
# find the Sum
print("Sum = ", calculateSum(n))
 
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to find sum of first n terms
using System;
class gfg
{
    // Function to calculate the sum
    public void calculateSum(int n)
    {
        double r = (n * (n + 1) / 2 +
                Math.Pow((n * (n + 1) / 2), 2));
        Console.WriteLine("Sum = " + r);
    }
 
    // Driver code
    public static int Main()
    {
        gfg g = new gfg();
 
        // number of terms to be
        // included in the sum
        int n = 3;
         
        // find the Sum
        g.calculateSum(n);
        Console.Read();
        return 0;
    }
}

Javascript

 
// Javascript program to find sum of first n terms
 
// Function to calculate the sum
function calculateSum(n)
{
 
    return n * (n + 1) / 2
        + Math.pow((n * (n + 1) / 2), 2);
}
 
// Driver code
 
    // number of terms to be
    // included in the sum
    let n = 3;
 
    // find the Sum
    document.write("Sum = " + calculateSum(n));
 
 
// This code is contributed by Mayank Tyagi
 

输出：

Sum = 42