Given a series:

S

_{n}= 1*3 + 3*5 + 5*7 + …

It is required to find the sum of first n terms of this series represented by S_{n}, where n is given taken input.**Examples**:

Input: n = 2Output: S<sub>n</sub> = 18Explanation: The sum of first 2 terms of Series is 1*3 + 3*5 = 3 + 15 = 28Input: n = 4Output: S<sub>n</sub> = 116Explanation: The sum of first 4 terms of Series is 1*3 + 3*5 + 5*7 + 7*9 = 3 + 15 + 35 + 63 = 116

Let, the n-th term be denoted by t_{n}.

This problem can easily be solved by observing that the nth term can be founded by following method:

t

_{n}= (n-th term of (1, 3, 5, … ) )*(nth term of (3, 5, 7, ….))

Now, n-th term of series 1, 3, 5 is given by **2*n-1**

and, the n-th term of series 3, 5, 7 is given by **2*n+1**

Putting these two values in t_{n}:

t

_{n}= (2*n-1)*(2*n+1) = 4*n*n-1

Now, the sum of first n terms will be given by :

S

_{n}= ∑(4*n*n – 1)

=∑4*{n*n}-∑(1)

Now, it is known that the sum of first n terms of series n*n (1, 4, 9, …) is given by: n*(n+1)*(2*n+1)/6

And sum of n number of 1’s is n itself.

Now, putting values in S_{n}:

S

_{n}= 4*n*(n+1)*(2*n+1)/6 – n

= n*(4*n*n + 6*n – 1)/3

Now, S_{n} value can be easily found by putting the desired value of n.

Below is the implementation of the above approach:

## C++

`// C++ program to find sum of first n terms` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `calculateSum(` `int` `n)` `{` ` ` `// Sn = n*(4*n*n + 6*n - 1)/3` ` ` `return` `(n * (4 * n * n + 6 * n - 1) / 3);` `}` `int` `main()` `{` ` ` `// number of terms to be included in the sum` ` ` `int` `n = 4;` ` ` `// find the Sn` ` ` `cout << ` `"Sum = "` `<< calculateSum(n);` ` ` `return` `0;` `}` |

## Java

`// Java program to find sum` `// of first n terms` `class` `GFG` `{` ` ` `static` `int` `calculateSum(` `int` `n)` ` ` `{` ` ` `// Sn = n*(4*n*n + 6*n - 1)/3` ` ` `return` `(n * (` `4` `* n * n +` ` ` `6` `* n - ` `1` `) / ` `3` `);` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `// number of terms to be` ` ` `// included in the sum` ` ` `int` `n = ` `4` `;` ` ` ` ` `// find the Sn` ` ` `System.out.println(` `"Sum = "` `+` ` ` `calculateSum(n));` ` ` `}` `}` `// This code is contributed by Bilal` |

## Python

`# Python program to find sum` `# of first n terms` `def` `calculateSum(n):` ` ` ` ` `# Sn = n*(4*n*n + 6*n - 1)/3` ` ` `return` `(n ` `*` `(` `4` `*` `n ` `*` `n ` `+` ` ` `6` `*` `n ` `-` `1` `) ` `/` `3` `);` `# Driver Code` `# number of terms to be` `# included in the sum` `n ` `=` `4` `# find the Sn` `print` `(` `"Sum ="` `,calculateSum(n))` `# This code is contributed by Bilal` |

## C#

`// C# program to find sum` `// of first n terms` `using` `System;` `class` `GFG` `{` `static` `int` `calculateSum(` `int` `n)` `{` ` ` `// Sn = n*(4*n*n + 6*n - 1)/3` ` ` `return` `(n * (4 * n * n +` ` ` `6 * n - 1) / 3);` `}` `// Driver code` `static` `public` `void` `Main ()` `{` ` ` `// number of terms to be` ` ` `// included in the sum` ` ` `int` `n = 4;` ` ` `// find the Sn` ` ` `Console.WriteLine(` `"Sum = "` `+` ` ` `calculateSum(n));` `}` `}` `// This code is contributed` `// by mahadev` |

## PHP

`<?php` `// PHP program to find sum` `// of first n terms` `function` `calculateSum(` `$n` `)` `{` ` ` `// Sn = n*(4*n*n + 6*n - 1)/3` ` ` `return` `(` `$n` `* (4 * ` `$n` `* ` `$n` `+` ` ` `6 * ` `$n` `- 1) / 3);` `}` `// number of terms to be` `// included in the sum` `$n` `= 4;` `// find the Sn` `echo` `"Sum = "` `. calculateSum(` `$n` `);` `// This code is contributed` `// by ChitraNayal` `?>` |

## Javascript

`<script>` `// Javascript program to find sum` `// of first n terms` ` ` `function` `calculateSum( n) {` ` ` `// Sn = n*(4*n*n + 6*n - 1)/3` ` ` `return` `(n * (4 * n * n + 6 * n - 1) / 3);` ` ` `}` ` ` `// Driver Code` ` ` ` ` `// number of terms to be` ` ` `// included in the sum` ` ` `let n = 4;` ` ` `// find the Sn` ` ` `document.write(` `"Sum = "` `+ calculateSum(n));` ` ` `// This code contributed by Princi Singh` `</script>` |

**Output:**

Sum = 116

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