# Program to find Sum of the series 1*3 + 3*5 + ….

• Difficulty Level : Easy
• Last Updated : 01 Nov, 2021

Given a series:

Sn = 1*3 + 3*5 + 5*7 + …

It is required to find the sum of first n terms of this series represented by Sn, where n is given taken input.
Examples

```Input : n = 2
Output : S<sub>n</sub> = 18
Explanation:
The sum of first 2 terms of Series is
1*3 + 3*5
= 3 + 15
= 28

Input : n = 4
Output : S<sub>n</sub> = 116
Explanation:
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 tn
This problem can easily be solved by observing that the nth term can be founded by following method:

tn = (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 tn:

tn = (2*n-1)*(2*n+1) = 4*n*n-1

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

Sn = ∑(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 Sn:

Sn = 4*n*(n+1)*(2*n+1)/6 – n
= n*(4*n*n + 6*n – 1)/3

Now, Sn 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 ` `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

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## Javascript

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Output:

`Sum = 116`

Time Complexity: O(1)

Auxiliary Space: O(1)

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