Related Articles
Sum of the series 1, 2, 4, 3, 5, 7, 9, 6, 8, 10, 11, 13.. till N-th term
• Last Updated : 10 May, 2019

Given a series of numbers 1, 2, 4, 3, 5, 7, 9, 6, 8, 10, 11, 13… The task is to find the sum of all the numbers in series till N-th number.

Examples:

Input: N = 4
Output: 10
1 + 2 + 4 + 3 = 10

Input: N = 10
Output: 55

Approach: The series is basically 20 odd numbers, 21 even numbers, 22 even numbers…. The sum of first N odd numbers is N * N and sum of first N even numbers is (N * (N+1)). Calculate the summation for 2i odd or even numbers and keep adding them to the sum.

Iterate for every power of 2, till the number of iterations exceeds N, and keep adding the respective summation of odd or even numbers according to the parity. For every segment the sum of the segment will be, (current sum of X odd/even numbers – previous sum of Y odd/even numbers), where X is the total sum of odd/even numbers till this segment and Y is the summation of odd/even numbers till the previous when odd/even numbers occurred.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement ` `// the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the ` `// sum of first N odd numbers ` `int` `sumodd(``int` `n) ` `{ ` `    ``return` `(n * n); ` `} ` ` `  `// Function to find the ` `// sum of first N even numbers ` `int` `sumeven(``int` `n) ` `{ ` `    ``return` `(n * (n + 1)); ` `} ` ` `  `// Function to overall ` `// find the sum of series ` `int` `findSum(``int` `num) ` `{ ` ` `  `    ``// Initiall odd numbers ` `    ``int` `sumo = 0; ` ` `  `    ``// Initial even numbers ` `    ``int` `sume = 0; ` ` `  `    ``// First power of 2 ` `    ``int` `x = 1; ` ` `  `    ``// Check for parity ` `    ``// for odd/even ` `    ``int` `cur = 0; ` ` `  `    ``// Counts the sum ` `    ``int` `ans = 0; ` `    ``while` `(num > 0) { ` ` `  `        ``// Get the minimum ` `        ``// out of remaining num ` `        ``// or power of 2 ` `        ``int` `inc = min(x, num); ` ` `  `        ``// Decrease that much numbers ` `        ``// from num ` `        ``num -= inc; ` ` `  `        ``// If the segment has odd numbers ` `        ``if` `(cur == 0) { ` ` `  `            ``// Summate the odd numbers ` `            ``// By exclusion ` `            ``ans = ans + sumodd(sumo + inc) - sumodd(sumo); ` ` `  `            ``// Increase number of odd numbers ` `            ``sumo += inc; ` `        ``} ` `        ``// If the segment has evn numbers ` `        ``else` `{ ` ` `  `            ``// Summate the even numbers ` `            ``// By exclusion ` `            ``ans = ans + sumeven(sume + inc) - sumeven(sume); ` ` `  `            ``// Increase number of even numbers ` `            ``sume += inc; ` `        ``} ` ` `  `        ``// Next set of numbers ` `        ``x *= 2; ` ` `  `        ``// Change parity for odd/even ` `        ``cur ^= 1; ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 4; ` `    ``cout << findSum(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to implement ` `// the above approach ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to find the ` `    ``// sum of first N odd numbers ` `    ``static` `int` `sumodd(``int` `n) ` `    ``{ ` `        ``return` `(n * n); ` `    ``} ` ` `  `    ``// Function to find the ` `    ``// sum of first N even numbers ` `    ``static` `int` `sumeven(``int` `n) ` `    ``{ ` `        ``return` `(n * (n + ``1``)); ` `    ``} ` ` `  `    ``// Function to overall ` `    ``// find the sum of series ` `    ``static` `int` `findSum(``int` `num)  ` `    ``{ ` ` `  `        ``// Initiall odd numbers ` `        ``int` `sumo = ``0``; ` ` `  `        ``// Initial even numbers ` `        ``int` `sume = ``0``; ` ` `  `        ``// First power of 2 ` `        ``int` `x = ``1``; ` ` `  `        ``// Check for parity ` `        ``// for odd/even ` `        ``int` `cur = ``0``; ` ` `  `        ``// Counts the sum ` `        ``int` `ans = ``0``; ` `        ``while` `(num > ``0``)  ` `        ``{ ` ` `  `            ``// Get the minimum ` `            ``// out of remaining num ` `            ``// or power of 2 ` `            ``int` `inc = Math.min(x, num); ` ` `  `            ``// Decrease that much numbers ` `            ``// from num ` `            ``num -= inc; ` ` `  `            ``// If the segment has odd numbers ` `            ``if` `(cur == ``0``) ` `            ``{ ` ` `  `                ``// Summate the odd numbers ` `                ``// By exclusion ` `                ``ans = ans + sumodd(sumo + inc) - sumodd(sumo); ` ` `  `                ``// Increase number of odd numbers ` `                ``sumo += inc; ` `            ``}  ` `             `  `            ``// If the segment has evn numbers ` `            ``else` `            ``{ ` ` `  `                ``// Summate the even numbers ` `                ``// By exclusion ` `                ``ans = ans + sumeven(sume + inc) - sumeven(sume); ` ` `  `                ``// Increase number of even numbers ` `                ``sume += inc; ` `            ``} ` ` `  `            ``// Next set of numbers ` `            ``x *= ``2``; ` ` `  `            ``// Change parity for odd/even ` `            ``cur ^= ``1``; ` `        ``} ` ` `  `        ``return` `ans; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `n = ``4``; ` `        ``System.out.println(findSum(n)); ` ` `  `    ``} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## Python

 `# Python3 program to implement ` `# the above approach ` ` `  `# Function to find the ` `# sum of first N odd numbers ` `def` `sumodd(n): ` ` `  `    ``return` `(n ``*` `n) ` ` `  `# Function to find the ` `# sum of first N even numbers ` `def` `sumeven(n): ` ` `  `    ``return` `(n ``*` `(n ``+` `1``)) ` ` `  ` `  `# Function to overall ` `# find the sum of series ` `def` `findSum(num): ` ` `  ` `  `    ``# Initiall odd numbers ` `    ``sumo ``=` `0` ` `  `    ``# Initial even numbers ` `    ``sume ``=` `0` ` `  `    ``# First power of 2 ` `    ``x ``=` `1` ` `  `    ``# Check for parity ` `    ``# for odd/even ` `    ``cur ``=` `0` ` `  `    ``# Counts the sum ` `    ``ans ``=` `0` `    ``while` `(num > ``0``): ` ` `  `        ``# Get the minimum ` `        ``# out of remaining num ` `        ``# or power of 2 ` `        ``inc ``=` `min``(x, num) ` ` `  `        ``# Decrease that much numbers ` `        ``# from num ` `        ``num ``-``=` `inc ` ` `  `        ``# If the segment has odd numbers ` `        ``if` `(cur ``=``=` `0``): ` ` `  `            ``# Summate the odd numbers ` `            ``# By exclusion ` `            ``ans ``=` `ans ``+` `sumodd(sumo ``+` `inc) ``-` `sumodd(sumo) ` ` `  `            ``# Increase number of odd numbers ` `            ``sumo ``+``=` `inc ` `         `  `        ``# If the segment has evn numbers ` `        ``else``: ` ` `  `            ``# Summate the even numbers ` `            ``# By exclusion ` `            ``ans ``=` `ans ``+` `sumeven(sume ``+` `inc) ``-` `sumeven(sume) ` ` `  `            ``# Increase number of even numbers ` `            ``sume ``+``=` `inc ` `         `  ` `  `        ``# Next set of numbers ` `        ``x ``*``=` `2` ` `  `        ``# Change parity for odd/even ` `        ``cur ^``=` `1` `     `  `    ``return` `ans ` ` `  `# Driver code ` `n ``=` `4` `print``(findSum(n)) ` ` `  `# This code is contributed by mohit kumar `

## C#

 `// C# program to implement ` `// the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to find the ` `    ``// sum of first N odd numbers ` `    ``static` `int` `sumodd(``int` `n) ` `    ``{ ` `        ``return` `(n * n); ` `    ``} ` ` `  `    ``// Function to find the ` `    ``// sum of first N even numbers ` `    ``static` `int` `sumeven(``int` `n) ` `    ``{ ` `        ``return` `(n * (n + 1)); ` `    ``} ` ` `  `    ``// Function to overall ` `    ``// find the sum of series ` `    ``static` `int` `findSum(``int` `num)  ` `    ``{ ` ` `  `        ``// Initiall odd numbers ` `        ``int` `sumo = 0; ` ` `  `        ``// Initial even numbers ` `        ``int` `sume = 0; ` ` `  `        ``// First power of 2 ` `        ``int` `x = 1; ` ` `  `        ``// Check for parity ` `        ``// for odd/even ` `        ``int` `cur = 0; ` ` `  `        ``// Counts the sum ` `        ``int` `ans = 0; ` `        ``while` `(num > 0)  ` `        ``{ ` ` `  `            ``// Get the minimum ` `            ``// out of remaining num ` `            ``// or power of 2 ` `            ``int` `inc = Math.Min(x, num); ` ` `  `            ``// Decrease that much numbers ` `            ``// from num ` `            ``num -= inc; ` ` `  `            ``// If the segment has odd numbers ` `            ``if` `(cur == 0) ` `            ``{ ` ` `  `                ``// Summate the odd numbers ` `                ``// By exclusion ` `                ``ans = ans + sumodd(sumo + inc) - sumodd(sumo); ` ` `  `                ``// Increase number of odd numbers ` `                ``sumo += inc; ` `            ``}  ` `             `  `            ``// If the segment has evn numbers ` `            ``else` `            ``{ ` ` `  `                ``// Summate the even numbers ` `                ``// By exclusion ` `                ``ans = ans + sumeven(sume + inc) - sumeven(sume); ` ` `  `                ``// Increase number of even numbers ` `                ``sume += inc; ` `            ``} ` ` `  `            ``// Next set of numbers ` `            ``x *= 2; ` ` `  `            ``// Change parity for odd/even ` `            ``cur ^= 1; ` `        ``} ` ` `  `        ``return` `ans; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``int` `n = 4; ` `        ``Console.WriteLine(findSum(n)); ` ` `  `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## PHP

 ` 0)  ` `    ``{ ` ` `  `        ``// Get the minimum ` `        ``// out of remaining num ` `        ``// or power of 2 ` `        ``\$inc` `= min(``\$x``, ``\$num``); ` ` `  `        ``// Decrease that much numbers ` `        ``// from num ` `        ``\$num` `-= ``\$inc``; ` ` `  `        ``// If the segment has odd numbers ` `        ``if` `(``\$cur` `== 0) ` `        ``{ ` ` `  `            ``// Summate the odd numbers ` `            ``// By exclusion ` `            ``\$ans` `= ``\$ans` `+ sumodd(``\$sumo` `+ ``\$inc``) -  ` `                          ``sumodd(``\$sumo``); ` ` `  `            ``// Increase number of odd numbers ` `            ``\$sumo` `+= ``\$inc``; ` `        ``} ` `         `  `        ``// If the segment has evn numbers ` `        ``else`  `        ``{ ` ` `  `            ``// Summate the even numbers ` `            ``// By exclusion ` `            ``\$ans` `= ``\$ans` `+ sumeven(``\$sume` `+ ``\$inc``) -  ` `                          ``sumeven(``\$sume``); ` ` `  `            ``// Increase number of even numbers ` `            ``\$sume` `+= ``\$inc``; ` `        ``} ` ` `  `        ``// Next set of numbers ` `        ``\$x` `*= 2; ` ` `  `        ``// Change parity for odd/even ` `        ``\$cur` `^= 1; ` `    ``} ` ` `  `    ``return` `\$ans``; ` `} ` ` `  `// Driver code ` `\$n` `= 4; ` `echo` `findSum(``\$n``); ` ` `  `// This code contributed by princiraj1992 ` `?> `

Output:

```10
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :