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

  • Difficulty Level : Easy
  • Last Updated : 25 Mar, 2021

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), … 
 

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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 Nth term of the series be Tn. Sum of the series can be easily found by observing the Nth term of the series: 
 

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

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

Sn = ΣTn 
Sn = Σ[4n3 + 12n2 + 11n + 3] 
Sn = (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

 




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