# Sum of first N natural numbers with all powers of 2 added twice

Given an integer **N**, the task is to calculate the sum of first N natural numbers adding all powers of 2 twice to the sum.**Examples:**

Input:N = 4Output:17Explanation:

Sum =2+4+3+8= 17

Since 1, 2 and 4 are 2^{ 0}, 2^{ 1}and 2^{ 2}respectively, they are added twice to the sum.Input:N = 5Output:22Explanation:

The sum is equal to2+4+3+8+5 = 22,

because 1, 2 and 4 are 2^{ 0}, 2^{ 1}and 2^{ 2}respectively.Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the

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**Naive Approach:**

The simplest approach to solve this problem is to iterate upto **N**, and keep calculating the sum by adding every number once except the powers of **2**, which needs to be added twice. **Time Complexity:** O(N) **Auxiliary Space:** O(1)**Efficient Approach:**

Follow the steps below to optimize the above approach:

- Calculate sum of first N natural numbers by the formula
**(N * (N + 1)) / 2**. - Now, all powers of 2 needs to be added once more. Sum of all powers of 2 up to N can be calculated as
**2**.^{ log}_{2}^{(N) + 1}– 1

- Hence, the required sum is:

(N * (N + 1)) / 2 + 2

^{ log}_{2}^{(N) + 1}– 1

Below is the implementation of the above approach:

## C++

`// C++ program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to raise N to the` `// power P and return the value` `double` `power(` `int` `N, ` `int` `P)` `{` ` ` `return` `pow` `(N, P);` `}` `// Function to calculate the` `// log base 2 of an integer` `int` `Log2(` `int` `N)` `{` ` ` `// Calculate log2(N) indirectly` ` ` `// using log() method` ` ` `int` `result = (` `int` `)(` `log` `(N) / ` `log` `(2));` ` ` `return` `result;` `}` `// Function to calculate and` `// return the required sum` `double` `specialSum(` `int` `n)` `{` ` ` `// Sum of first N natural` ` ` `// numbers` ` ` `double` `sum = n * (n + 1) / 2;` ` ` `// Sum of all powers of 2` ` ` `// up to N` ` ` `int` `a = Log2(n);` ` ` `sum = sum + power(2, a + 1) - 1;` ` ` `return` `sum;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 4;` ` ` `cout << (specialSum(n)) << endl;` ` ` `return` `0;` `}` `// This code is contributed by divyeshrabadiya07` |

## Java

`// Java program to implement` `// the above approach` `import` `java.util.*;` `import` `java.lang.Math;` `class` `GFG {` ` ` `// Function to raise N to the` ` ` `// power P and return the value` ` ` `static` `double` `power(` `int` `N, ` `int` `P)` ` ` `{` ` ` `return` `Math.pow(N, P);` ` ` `}` ` ` `// Function to calculate the` ` ` `// log base 2 of an integer` ` ` `public` `static` `int` `log2(` `int` `N)` ` ` `{` ` ` `// Calculate log2(N) indirectly` ` ` `// using log() method` ` ` `int` `result = (` `int` `)(Math.log(N)` ` ` `/ Math.log(` `2` `));` ` ` `return` `result;` ` ` `}` ` ` `// Function to calculate and` ` ` `// return the required sum` ` ` `static` `double` `specialSum(` `int` `n)` ` ` `{` ` ` `// Sum of first N natural` ` ` `// numbers` ` ` `double` `sum = n * (n + ` `1` `) / ` `2` `;` ` ` `// Sum of all powers of 2` ` ` `// up to N` ` ` `int` `a = log2(n);` ` ` `sum = sum + power(` `2` `, a + ` `1` `) - ` `1` `;` ` ` `return` `sum;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `n = ` `4` `;` ` ` `System.out.println(specialSum(n));` ` ` `}` `}` |

## Python3

`# Python3 program to implement` `# the above approach` `import` `math` `# Function to raise N to the` `# power P and return the value` `def` `power(N, P):` ` ` ` ` `return` `math.` `pow` `(N, P)` `# Function to calculate the` `# log base 2 of an integer` `def` `Log2(N):` ` ` ` ` `# Calculate log2(N) indirectly` ` ` `# using log() method` ` ` `result ` `=` `(math.log(N) ` `/` `/` `math.log(` `2` `))` ` ` ` ` `return` `result` `# Function to calculate and` `# return the required sum` `def` `specialSum(n):` ` ` ` ` `# Sum of first N natural` ` ` `# numbers` ` ` `sum` `=` `n ` `*` `(n ` `+` `1` `) ` `/` `/` `2` ` ` `# Sum of all powers of 2` ` ` `# up to N` ` ` `a ` `=` `Log2(n)` ` ` `sum` `=` `sum` `+` `power(` `2` `, a ` `+` `1` `) ` `-` `1` ` ` ` ` `return` `sum` ` ` `# Driver code ` `if` `__name__` `=` `=` `"__main__"` `:` ` ` ` ` `n ` `=` `4` ` ` `print` `(specialSum(n))` `# This code is contributed by rutvik_56` |

## C#

`// C# program to implement` `// the above approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to raise N to the` ` ` `// power P and return the value` ` ` `static` `double` `power(` `int` `N, ` `int` `P)` ` ` `{` ` ` `return` `Math.Pow(N, P);` ` ` `}` ` ` `// Function to calculate the` ` ` `// log base 2 of an integer` ` ` `public` `static` `int` `log2(` `int` `N)` ` ` `{` ` ` `// Calculate log2(N) indirectly` ` ` `// using log() method` ` ` `int` `result = (` `int` `)(Math.Log(N) /` ` ` `Math.Log(2));` ` ` `return` `result;` ` ` `}` ` ` `// Function to calculate and` ` ` `// return the required sum` ` ` `static` `double` `specialSum(` `int` `n)` ` ` `{` ` ` `// Sum of first N natural` ` ` `// numbers` ` ` `double` `sum = (` `double` `)(n) * (n + 1) / 2;` ` ` `// Sum of all powers of 2` ` ` `// up to N` ` ` `int` `a = log2(n);` ` ` `sum = (sum) + power(2, a + 1) - 1;` ` ` `return` `sum;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(` `string` `[] args)` ` ` `{` ` ` `int` `n = 4;` ` ` `Console.Write(specialSum(n));` ` ` `}` `}` `// This code is contributed by Ritik Bansal` |

## Javascript

`<script>` `// Javascript program to implement` `// the above approach` `// Function to raise N to the` `// power P and return the value` `function` `power(N, P)` `{` ` ` `return` `Math.pow(N, P);` `}` `// Function to calculate the` `// log base 2 of an integer` `function` `Log2(N)` `{` ` ` `// Calculate log2(N) indirectly` ` ` `// using log() method` ` ` `let result = (Math.floor(Math.log(N) / Math.log(2)));` ` ` `return` `result;` `}` `// Function to calculate and` `// return the required sum` `function` `specialSum(n)` `{` ` ` `// Sum of first N natural` ` ` `// numbers` ` ` `let sum = n * (n + 1) / 2;` ` ` `// Sum of all powers of 2` ` ` `// up to N` ` ` `let a = Log2(n);` ` ` `sum = sum + power(2, a + 1) - 1;` ` ` `return` `sum;` `}` `// Driver Code` ` ` `let n = 4;` ` ` `document.write(specialSum(n) + ` `"<br>"` `);` `// This code is contributed by Mayank Tyagi` `</script>` |

**Output:**

17.0

**Time Complexity:** O(log_{2}(N)) **Auxiliary Space:** O(1)