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

A **simple solution **is to iterate through all odd numbers.

## C++

`// A naive CPP program to find sum of` `// first n odd numbers` `#include <iostream>` `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

`<?php` `// A naive PHP program to find sum of` `// first n odd numbers` ` ` `// Returns the sum of first` `// n odd numbers` `function` `oddSum(` `$n` `)` `{` ` ` `$sum` `= 0; ` `$curr` `= 1;` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `{` ` ` `$sum` `+= ` `$curr` `;` ` ` `$curr` `+= 2;` ` ` `}` ` ` `return` `$sum` `;` `}` ` ` `// Driver Code` `$n` `= 20;` `echo` `" Sum of first "` `, ` `$n` ` ` `, ` `" Odd Numbers is: "` `, oddSum(` `$n` `);` ` ` `// This code is contributed by vt_m.` `?>` |

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) = n

^{2}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 <iostream>` `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

`<?php` `// Efficient program to find sum of` `// first n odd numbers` ` ` `// Returns the sum of first` `// n odd numbers` `function` `oddSum(` `$n` `)` `{` ` ` `return` `(` `$n` `* ` `$n` `);` `}` ` ` `// Driver Code` `$n` `= 20;` `echo` `" Sum of first "` `, ` `$n` `,` ` ` `" Odd Numbers is: "` `, oddSum(` `$n` `);` ` ` `// This code is contributed by vt_m.` `?>` |

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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