求出系列2×3 + 4×4 + 6×5 + 8×6 +的前N个项的总和。

给定整数N。任务是找到高达给出一系列的N项之和：

2×3 + 4×4 + 6×5 + 8×6 + … + upto n terms 
 

例子：

Input : N = 5
Output : Sum = 170

Input : N = 10
Output : Sum = 990

令级数的第N个为t _N。
t ₁ = 2×3 =(2×1)(1 + 2)
t ₂ = 4×4 =(2×2)(2 + 2)
t ₃ = 6×5 =(2×3)(3 + 2)
t ₄ = 8×6 =(2×4)(4 + 2)
。
。
。
t _N =(2×N)(N + 2)
该系列的n个项之和，

S _n = t ₁ + t ₂ + … + t _n = $\;\sum_{k=1}^{n} t_{k}$ = $\;\sum_{k=1}^{n} 2k(k+2)$ = $\;\sum_{k=1}^{n} (2k^{2}+4k)$ = $\;\sum_{k=1}^{n} 2k^{2} + \;\sum_{k=1}^{n} 4k$ = $2\frac{n(n+1)(2n+1)}{6} + 4\frac{n(n+1)}{2}$ = $n(n+1)[ \frac{2n+1}{3} +2]$ = $\frac{n(n+1)(2n+7)}{3}$

下面是上述方法的实现：

C++

// C++ program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
#include
using namespace std;
 
// calculate sum upto N term of series
void Sum_upto_nth_Term(int n)
{
    int r = n * (n + 1) *
                (2 * n + 7) / 3;
    cout << r;
}
 
// Driver code
int main()
{
    int N = 5;
    Sum_upto_nth_Term(N) ;
    return 0;
}

Java

// Java program to find sum upto
// N term of the series:
 
import java.io.*;
 
class GFG {
// calculate sum upto N term of series
static void Sum_upto_nth_Term(int n)
{
    int r = n * (n + 1) *
                (2 * n + 7) / 3;
    System.out.println(r);
}
 
// Driver code
    public static void main (String[] args) {
    int N = 5;
    Sum_upto_nth_Term(N);
    }
}

Python3

# Python program to find sum upto N term of the series:
# 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
 
# calculate sum upto N term of series
def Sum_upto_nth_Term(n):
    return n * (n + 1) * (2 * n + 7) // 3
 
# Driver code
N = 5
print(Sum_upto_nth_Term(N))

C#

// C# program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
 
using System;
 
class GFG
{
// calculate sum upto N term of series
static void Sum_upto_nth_Term(int n)
{
    int r = n * (n + 1) *
                (2 * n + 7) / 3;
    Console.Write(r);
}
 
// Driver code
public static void Main()
{
    int N = 5;
    Sum_upto_nth_Term(N);
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Javascript

 
// Javascript program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
 
// calculate sum upto N term of series
function Sum_upto_nth_Term(n)
{
    let r = n * (n + 1) *
                (2 * n + 7) / 3;
    document.write(r);
}
 
// Driver code
 
    let N = 5;
    Sum_upto_nth_Term(N) ;
 
 
// This code is contributed by Mayank Tyagi
 

输出：

170