Find the sum of first N terms of the series 2*3*5, 3*5*7, 4*7*9, …

Given an integer N, the task is to find the sum of first N terms of the series: 

(2 * 3 * 5), (3 * 5 * 7), (4 * 7 * 9), … 

Examples:

Input: N = 3 
Output: 387 
S₃ = (2 * 3 * 5) + (3 * 5 * 7) + (4 * 7 * 9) = 30 + 105 + 252 = 387
Input: N = 5 
Output: 1740 

Approach: Let the N^th term of the series be T_n. Sum of the series can be easily found by observing the N^th term of the series: 

T_n = {n^th term of 2, 3, 4, …} * {n^th term of 3, 5, 7, …} * {n^th term of 5, 7, 9, …} 
T_n = (n + 1) * (2 * n + 1) * (2* n + 3) 
T_n = 4n³ + 12n² + 11n + 3

Sum(S_n) of first n terms can be found by

S_n = ΣT_n
S_n = Σ[4n³ + 12n² + 11n + 3] 
S_n = (n / 2) * [2n³ + 12n² + 25n + 21]

Below is the implementation of the above approach:

C++

// C++ program to find sum of the
// first n terms of the given series
#include <bits/stdc++.h>
using namespace std;

// Function to return the sum of the
// first n terms of the given series
int calSum(int n)
{
    // As described in the approach
    return (n * (2 * n * n * n + 12 * n * n + 25 * n + 21)) / 2;
}

// Driver code
int main()
{
    int n = 3;
    cout << calSum(n);
    return 0;
}

Java

// Java program to find sum of the
// first n terms of the given series
class GFG {

    // Function to return the sum of the
    // first n terms of the given series
    static int calSum(int n)
    {

        // As described in the approach
        return (n * (2 * n * n * n + 12 * n * n + 25 * n + 21)) / 2;
    }

    // Driver Code
    public static void main(String args[])
    {
        int n = 3;
        System.out.println(calSum(n));
    }
}

Python

# C++ program to find sum of the
# first n terms of the given series

# Function to return the sum of the
# first n terms of the given series
def calSum(n):

    # As described in the approach
    return (n*(2 * n*n * n + 12 * n*n + 25 * n + 21))/2;

# Driver Code
n = 3
print(calSum(n))

C#

// C# program to find sum of the
// first n terms of the given series
using System;

class GFG {

    // Function to return the sum of the
    // first n terms of the given series
    static int calSum(int n)
    {

        // As described in the approach
        return (n * (2 * n * n * n + 12 * n * n + 25 * n + 21)) / 2;
    }

    // Driver code
    static public void Main()
    {
        int n = 3;
        Console.WriteLine(calSum(n));
    }
}

PHP

<?php
// PHP script to find sum of the
// first n terms of the given series

// Function to return the sum of the
// first n terms of the given series
function calculateSum($n)
{

    // As described in the approach
    return ($n*(2*$n*$n*$n+12*$n*$n+25*$n+21))/2;
}

// Driver code
$n = 3;
echo calculateSum($n);

?>

Javascript

<script>
// Javascript program to find sum of the
// first n terms of the given series

    // Function to return the sum of the
    // first n terms of the given series
    function calSum( n) {

        // As described in the approach
        return (n * (2 * n * n * n + 12 * n * n + 25 * n + 21)) / 2;
    }

    // Driver Code
    let n = 3;
    document.write(calSum(n));

// This code is contributed by 29AjayKumar
</script>

Output:

387

Time Complexity: O(1)

Auxiliary Space: O(1), since no extra space has been taken.