# 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 + …

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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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