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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 : 27 Aug, 2022

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 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 = Î£Tn
Sn = Î£[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 ` `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

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## Javascript

 ``

Output:

`387`

Time Complexity: O(1)

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

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