# Sum of first n odd numbers in O(1) Complexity

Given the sequence of odd numbers
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, ….
Find the sum of first n odd numbers

Examples:

```Input : n = 2
Output : 4
Sum of first two odd numbers is 1 + 3 = 4.

Input : 5
Output : 25
Sum of first 5 odd numbers is 1 + 3 + 5 +
7 + 9 = 25
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to iterate through all odd numbers.

## C++

 `// A naive CPP program to find sum of ` `// first n odd numbers ` `#include ` `using` `namespace` `std; ` ` `  `// Returns the sum of first ` `// n odd numbers ` `int` `oddSum(``int` `n) ` `{ ` `    ``int` `sum = 0, curr = 1; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``sum += curr; ` `        ``curr += 2; ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Driver function ` `int` `main() ` `{ ` `    ``int` `n = 20; ` `    ``cout << ``" Sum of first "` `<< n ` `         ``<< ``" Odd Numbers is: "` `<< oddSum(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find sum of ` `// first n odd numbers ` `import` `java.util.*; ` ` `  `class` `Odd ` `{    ` `    ``// Returns the sum of first ` `    ``// n odd numbers ` `    ``public` `static` `int` `oddSum(``int` `n) ` `    ``{ ` `        ``int` `sum = ``0``, curr = ``1``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``sum += curr; ` `            ``curr += ``2``; ` `        ``} ` `        ``return` `sum; ` `    ``} ` `     `  `    ``// driver function ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``20``; ` `        ``System.out.println(``" Sum of first "``+ n ` `        ``+``" Odd Numbers is: "``+oddSum(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by rishabh_jain `

## Python3

 `# Python3 program to find sum ` `# of first n odd numbers ` ` `  `def` `oddSum(n) : ` `    ``sum` `=` `0` `    ``curr ``=` `1` `    ``i ``=` `0` `    ``while` `i < n: ` `        ``sum` `=` `sum` `+` `curr ` `        ``curr ``=` `curr ``+` `2` `        ``i ``=` `i ``+` `1` `    ``return` `sum` ` `  `# Driver Code ` `n ``=` `20` `print` `(``" Sum of first"` `, n, ``"Odd Numbers is: "``, ` `                                ``oddSum(n) )  ` ` `  `# This code is contributed by rishabh_jain `

## C#

 `// C# program to find sum of ` `// first n odd numbers ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Returns the sum of first ` `    ``// n odd numbers ` `    ``public` `static` `int` `oddSum(``int` `n) ` `    ``{ ` `        ``int` `sum = 0, curr = 1; ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``sum += curr; ` `            ``curr += 2; ` `        ``} ` `         `  `        ``return` `sum; ` `    ``} ` ` `  `    ``// driver function ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 20; ` `        ``Console.WriteLine(``" Sum of first "` `+ n ` `            ``+ ``" Odd Numbers is: "` `+ oddSum(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

`Sum of first 20 odd numbers is 400`

Time Complexity: O(n)
Auxiliary Space : O(1)

An efficient solution is to use direct formula. To find the sum of first n odd numbers we can apply odd number theorem, it states that the sum of first n odd numbers is equal to the square of n.

∑(2i – 1) = n2 where i varies from 1 to n

let n = 10, therefore sum of first 10 odd numbers is

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

if we apply odd number theorem:

sum of first 10 odd numbers = n * n = 10 * 10 = 100.

Below is the implementation of the above approach:

## C++

 `// Efficient program to find sum of ` `// first n odd numbers ` `#include ` `using` `namespace` `std; ` ` `  `// Returns the sum of first ` `// n odd numbers ` `int` `oddSum(``int` `n) ` `{ ` `    ``return` `(n * n); ` `} ` ` `  `// Driver function ` `int` `main() ` `{ ` `    ``int` `n = 20; ` `    ``cout << ``" Sum of first "` `<< n ` `         ``<< ``" Odd Numbers is: "` `<< oddSum(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find sum of ` `// first n odd numbers ` `import` `java.util.*; ` ` `  `class` `Odd ` `{    ` `    ``// Returns the sum of first ` `    ``// n odd numbers ` `    ``public` `static` `int` `oddSum(``int` `n) ` `    ``{ ` `        ``return` `(n * n); ` `    ``} ` `     `  `    ``// driver function ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``20``; ` `        ``System.out.println(``" Sum of first "``+ n ` `        ``+``" Odd Numbers is: "``+oddSum(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by rishabh_jain `

## Python3

 `# Python3 program to find sum ` `# of first n odd numbers ` ` `  `def` `oddSum(n) : ` `    ``return` `(n ``*` `n); ` ` `  `# Driver Code ` `n ``=` `20` `print` `(``" Sum of first"` `, n, ``"Odd Numbers is: "``, ` `                               ``oddSum(n) )  ` ` `  `# This code is contributed by rishabh_jain `

## C#

 `// C# program to find sum of ` `// first n odd numbers ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Returns the sum of first ` `    ``// n odd numbers ` `    ``public` `static` `int` `oddSum(``int` `n) ` `    ``{ ` `        ``return` `(n * n); ` `    ``} ` ` `  `    ``// driver function ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 20; ` `        ``Console.WriteLine(``" Sum of first "` `+ n ` `            ``+ ``" Odd Numbers is: "` `+ oddSum(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```Sum of first 20 odd numbers is 400
```

Time Complexity: O(1)
Auxiliary Space : O(1)

How does it work?
We can prove it using mathematical induction. We know it is true for n = 1 and n = 2 as sums are 1 and 4 (1 + 3) respectively.

```Let it be true for n = k-1.

Sum of first k odd numbers =
Sum of first k-1 odd numbers + k'th odd number
= (k-1)*(k-1) + (2k - 1)
= k*k```

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