# Sum of first n term of Series 3, 5, 9, 17, 33….

Given n, we need to find sum of first n terms of the series represented as Sn = 3 + 5 + 9 + 17 + 33 … upto n**Examples:**

Input : 2 Output : 8 3 + 5 = 8 Input : 5 Output : 67 3 + 5 + 9 + 17 + 33 = 67

Let, the nth term be denoted by tn.

This problem can easily be solved by splitting each term as follows :

Sn = 3 + 5 + 9 + 17 + 33……

Sn = (2+1) + (4+1) + (8+1) + (16+1) +……

Sn = (2+1) + (2*2+1) + (2*2*2+1) + (2*2*2*2+1) +……+ ((2*2*2..unto n times) + 1)

We observed that the nth term can be written in terms of powers of 2 and 1.

Hence, the sum of first n terms is given as follows:

Sn = (2+1) + (4+1) + (8+1) + (16+1) +……+ upto n terms

Sn = (1 + 1 + 1 + 1 + …unto n terms) + (2 + 4 + 8 + 16 + …upto nth power of 2)

In above formula,

2 + 4 + 8 + 16…. is a G.P.

It’s sum of first n terms is given by 2*(2^n-1)/(2-1) = 2^(n+1) – 2 (using G.P. formula)

Sn = n + 2*(2^n – 1)

Sn = 2^(n+1) + n -2

## C++

`// C++ program to find sum of first n terms` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `calculateSum(` `int` `n)` `{` ` ` `// Sn = n*(4*n*n + 6*n - 1)/3` ` ` `return` `(` `pow` `(2, n + 1) + n - 2);` `}` `// Driver code` `int` `main()` `{` ` ` `// number of terms to be included in sum` ` ` `int` `n = 4;` ` ` `// find the Sn` ` ` `cout << ` `"Sum = "` `<< calculateSum(n);` ` ` `return` `0;` `}` |

## Java

`// Java program to find` `// sum of first n terms` `import` `java.util.*;` `class` `GFG` `{` `static` `int` `calculateSum(` `int` `n)` `{` ` ` `// Sn = n*(4*n*n + 6*n - 1)/3` ` ` `return` `((` `int` `)Math.pow(` `2` `, n + ` `1` `) +` ` ` `n - ` `2` `);` `}` `// Driver Code` `public` `static` `void` `main(String args[])` `{` ` ` `// number of terms to` ` ` `// be included in sum` ` ` `int` `n = ` `4` `;` ` ` `// find the Sn` ` ` `System.out.println(` `"Sum = "` `+` ` ` `calculateSum(n));` `}` `}` `// This code is contributed` `// by Kirti_Mangal` |

## Python

`# Python program to find sum` `# of n terms of the series` `def` `calculateSum(n):` ` ` `return` `(` `2` `*` `*` `(n ` `+` `1` `) ` `+` `n ` `-` `2` `)` `# Driver Code` `# number of terms for the sum` `n ` `=` `4` `# find the Sn` `print` `(` `"Sum ="` `, calculateSum(n))` `# This code is contributed` `# by Surendra_Gangwar` |

## 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` `((` `int` `)Math.Pow(2, n + 1) +` ` ` `n - 2);` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` `// number of terms to` ` ` `// be included in sum` ` ` `int` `n = 4;` ` ` `// find the Sn` ` ` `Console.WriteLine(` `"Sum = "` `+` ` ` `calculateSum(n));` `}` `}` `// This code is contributed` `// by inder_verma..` |

## PHP

`<?php` `// PHP program to find sum` `// of first n terms` `function` `calculateSum( ` `$n` `)` `{` ` ` `// Sn = n*(4*n*n + 6*n - 1)/3` ` ` `return` `(pow(2, ` `$n` `+ 1) + ` `$n` `- 2);` `}` `// Driver code` `// number of terms to be` `// included in sum` `$n` `= 4;` `// find the Sn` `echo` `"Sum = "` `, calculateSum(` `$n` `);` `// This code is contributed` `// by inder_verma..` `?>` |

## Javascript

`<script>` `// Java script program to find` `// sum of first n terms` `function` `calculateSum( n)` `{` ` ` `// Sn = n*(4*n*n + 6*n - 1)/3` ` ` `return` `(Math.pow(2, n + 1) +` ` ` `n - 2);` `}` `// Driver Code` ` ` `// number of terms to` ` ` `// be included in sum` ` ` `let n = 4;` ` ` `// find the Sn` ` ` `document.write(` `"Sum = "` `+` ` ` `calculateSum(n));` `// This code is contributed by mohan pavan` `</script>` |

**Output:**

Sum = 34

**Time Complexity: **O(log n)

**Auxiliary Space: **O(log n)