# Find the sum of the series 2, 5, 13, 35, 97…

Given a series and a number n, the task is to find the sum of its first n terms. Below is the series:

2, 5, 13, 35, 97, …

**Examples:**

Input: N = 2 Output: 7 The sum of first 2 terms of Series is 2 + 5 = 7 Input: N = 4 Output: 55 The sum of first 4 terms of Series is 2 + 5 + 13 + 35 = 55

**Approach: **From this given series we find it is the sum of the Two **GP** series with common ratioes 2, 3 .

Sn = 2 + 5 + 13 + 35 + 97 … + upto nth term

Sn = (2^0 + 3^ 0) + (2^1 + 3^1) + (2^2 + 3^2) + (2^3 + 3^3)+ (2^4 + 3^4) …… + upto nth term

Sn = (2^0 + 2^1 + 2^2 + 2^3 + 2^4 … + upto nth term) + ( 3^0 + 3^1 + 3^2 + 3^3 …… + unto nth term )

Since, We know that the sum of n terms of the **GP** is given by the following formula:

**Below is the implementation of the above approach:**

## C++

`// C++ program for finding the sum` `// of first N terms of the series.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// CalculateSum function returns the final sum` `int` `calculateSum(` `int` `n)` `{` ` ` `// r1 and r2 are common ratios` ` ` `// of 1st and 2nd series` ` ` `int` `r1 = 2, r2 = 3;` ` ` `// a1 and a2 are common first terms` ` ` `// of 1st and 2nd series` ` ` `int` `a1 = 1, a2 = 1;` ` ` `return` `a1 * (` `pow` `(r1, n) - 1) / (r1 - 1)` ` ` `+ a2 * (` `pow` `(r2, n) - 1) / (r2 - 1);` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 4;` ` ` `// function calling for 4 terms` ` ` `cout << ` `"Sum = "` `<< calculateSum(n)` ` ` `<< endl;` ` ` `return` `0;` `}` |

## Java

`//Java program for finding the sum` `//of first N terms of the series.` `public` `class` `GFG {` ` ` `//CalculateSum function returns the final sum` ` ` `static` `int` `calculateSum(` `int` `n)` ` ` `{` ` ` `// r1 and r2 are common ratios` ` ` `// of 1st and 2nd series` ` ` `int` `r1 = ` `2` `, r2 = ` `3` `;` ` ` `// a1 and a2 are common first terms` ` ` `// of 1st and 2nd series` ` ` `int` `a1 = ` `1` `, a2 = ` `1` `;` ` ` `return` `(` `int` `)(a1 * (Math.pow(r1, n) - ` `1` `) / (r1 - ` `1` `)` ` ` `+ a2 * (Math.pow(r2, n) - ` `1` `) / (r2 - ` `1` `));` ` ` `}` ` ` `//Driver code` ` ` `public` `static` `void` `main(String[] args) {` ` ` ` ` `int` `n = ` `4` `;` ` ` `// function calling for 4 terms` ` ` `System.out.println(` `"Sum = "` `+calculateSum(n));` ` ` `}` `}` |

## Python 3

`# Python 3 program for finding the sum` `# of first N terms of the series.` `# from math import everything` `from` `math ` `import` `*` `# CalculateSum function returns the final sum` `def` `calculateSum(n) :` ` ` `# r1 and r2 are common ratios` ` ` `# of 1st and 2nd series` ` ` `r1, r2 ` `=` `2` `, ` `3` ` ` `# a1 and a2 are common first terms` ` ` `# of 1st and 2nd series` ` ` `a1, a2 ` `=` `1` `, ` `1` ` ` `return` `(a1 ` `*` `(` `pow` `(r1, n) ` `-` `1` `) ` `/` `/` `(r1 ` `-` `1` `)` ` ` `+` `a2 ` `*` `(` `pow` `(r2, n) ` `-` `1` `) ` `/` `/` `(r2 ` `-` `1` `))` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `n ` `=` `4` ` ` `# function calling for 4 terms` ` ` `print` `(` `"SUM = "` `,calculateSum(n))` `# This code is contributed by ANKITRAI1` |

## C#

`// C# program for finding the sum` `// of first N terms of the series.` `using` `System;` `class` `GFG` `{` `// CalculateSum function` `// returns the final sum` `static` `int` `calculateSum(` `int` `n)` `{` `// r1 and r2 are common ratios` `// of 1st and 2nd series` `int` `r1 = 2, r2 = 3;` `// a1 and a2 are common first` `// terms of 1st and 2nd series` `int` `a1 = 1, a2 = 1;` `return` `(` `int` `)(a1 * (Math.Pow(r1, n) - 1) / (r1 - 1) +` ` ` `a2 * (Math.Pow(r2, n) - 1) / (r2 - 1));` `}` `// Driver code` `static` `public` `void` `Main ()` `{` ` ` `int` `n = 4;` ` ` `// function calling for 4 terms` ` ` `Console.Write(` `"Sum = "` `+` ` ` `calculateSum(n));` `}` `}` `// This code is contributed by Raj` |

## PHP

`<?php` `// PHP program for finding the sum` `// of first N terms of the series.` `// CalculateSum function returns` `// the final sum` `function` `calculateSum(` `$n` `)` `{` ` ` `// r1 and r2 are common ratios` ` ` `// of 1st and 2nd series` ` ` `$r1` `= 2;` ` ` `$r2` `= 3;` ` ` `// a1 and a2 are common first ` ` ` `// terms of 1st and 2nd series` ` ` `$a1` `= 1;` ` ` `$a2` `= 1;` ` ` `return` `$a1` `* (pow(` `$r1` `, ` `$n` `) - 1) /` ` ` `(` `$r1` `- 1) + ` `$a2` `*` ` ` `(pow(` `$r2` `, ` `$n` `) - 1) /` ` ` `(` `$r2` `- 1);` `}` `// Driver code` `$n` `= 4;` `// function calling for 4 terms` `echo` `"Sum = "` `, calculateSum(` `$n` `);` `// This code is contributed by ash264` `?>` |

## Javascript

`<script>` `//javascript program for finding the sum` `//of first N terms of the series.` `//CalculateSum function returns the final sum` `function` `calculateSum(n)` `{` ` ` `// r1 and r2 are common ratios` ` ` `// of 1st and 2nd series` ` ` `var` `r1 = 2, r2 = 3;` ` ` `// a1 and a2 are common first terms` ` ` `// of 1st and 2nd series` ` ` `var` `a1 = 1, a2 = 1;` ` ` `return` `parseInt((a1 * (Math.pow(r1, n) - 1) / (r1 - 1)` ` ` `+ a2 * (Math.pow(r2, n) - 1) / (r2 - 1)));` `}` ` ` `//Driver code` ` ` ` ` `var` `n = 4;` ` ` `// function calling for 4 terms` ` ` `document.write(` `"Sum = "` `+calculateSum(n));` `// This code contributed by shikhasingrajput` `</script>` |

**Output:**

Sum = 55