Find the sum of first N terms of the series 2×3 + 4×4 + 6×5 + 8×6 + …

Given an integer N. The task is to find the sum upto N terms of the given series:

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

Examples:

Input : N = 5
Output : Sum = 170

Input : N = 10
Output : Sum = 990

Let the N-th term of the series be t_N. 
t₁ = 2 × 3 = (2 × 1)(1 + 2) 
t₂ = 4 × 4 = (2 × 2)(2 + 2) 
t₃ = 6 × 5 = (2 × 3)(3 + 2) 
t₄ = 8 × 6 = (2 × 4)(4 + 2) 
. 
. 
. 
t_N = (2 × N)(N + 2)
The sum of n terms of the series, 

 Sn = t1 + t2 +... + tn
    =======

Below is the implementation of above approach: 

C++

// C++ program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
#include<iostream>
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)

PHP

<?php
// PHP program to find sum upto
// N term of the series:
// 2 × 3 + 4 × 4 + 6 × 5 + 8 × 6 + ...
function Sum_upto_nth_Term($n)
{
    $r = $n * ($n + 1) *
              (2 * $n + 7) / 3;
    echo $r;
}

// Driver code
$N = 5;
Sum_upto_nth_Term($N);

// This code is contributed
// by Shashank_Sharma
?>

Javascript

<script>

// 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

</script>

Output:

170

Time Complexity: O(1), it is a constant.
Auxiliary Space: O(1), no extra space is required.