# 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 = 3Output:387

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

**Approach:** Let the **N ^{th}** term of the series be

**T**. Sum of the series can be easily found by observing the

_{n}**N**term of the series:

^{th}

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 + 1) * (2 * n + 1) * (2* n + 3)_{n}T_{n}= 4n^{3}+ 12n^{2}+ 11n + 3

Sum(**S _{n}**) of first n terms can be found by

S= Σ_{n}T_{n}S= Σ[4n_{n}^{3}+ 12n^{2}+ 11n + 3]S_{n}= (n / 2) * [2n^{3}+ 12n^{2}+ 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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