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

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

 ` `

Output:

```Sum = 116
```

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