# Sum of first n terms of a given series 3, 6, 11, …..

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

3, 6, 11, 20, ….

Examples:

Input: N = 2
Output: 9
The sum of first 2 terms of Series is
3 + 6 = 9

Input: N = 3
Output: 20
The sum of first 3 terms of Series is
3 + 6 + 11 = 20


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This problem can easily solved by observing that the nth term of the series :

Sn = 3 + 6 + 11 + 20 … + upto nth term
Sn = (1 + 2^1) + (2 + 2^2) + (3 + 2^3)+ (4 + 2^4) …… + upto nth term
Sn = (1 + 2 + 3 + 4 …. + upto nth term) + ( 2^1 + 2^2 + 2^3 …… + unto nth term )

We observe that Sn is a summation of two series: AP and GP
As we know the sum of first n terms of AP is given by And also the Sum of first n terms of G.P is given by Hence the total sum is given by sum of both AP and GP. Below is the implementation of above approach.

## C++

 // C++ program to find sum of first n terms  #include  using namespace std;     // Function to calculate the sum  int calculateSum(int n)  {      // starting number      int a1 = 1, a2 = 2;         // Common Ratio      int r = 2;         // Common difference      int d = 1;         return (n) * (2 * a1 + (n - 1) * d) / 2 + a2                  * (pow(r, n) - 1) / (r - 1);  }     // Driver code  int main()  {         // N th term to be find      int n = 5;         // find the Sn      cout << "Sum = " << calculateSum(n);         return 0;  }

## Java

 // Java program to find sum of first n terms     import java.io.*;     class GFG {     // Function to calculate the sum  static int calculateSum(int n)  {      // starting number      int a1 = 1, a2 = 2;         // Common Ratio      int r = 2;         // Common difference      int d = 1;         return (n) * (2 * a1 + (n - 1) * d) / 2 + a2               * (int)(Math.pow(r, n) - 1) / (r - 1);  }     // Driver code      public static void main (String[] args) {          // N th term to be find      int n = 5;         // find the Sn      System.out.print( "Sum = " + calculateSum(n));      }  }  // This code is contributed by inder_verma.

## Python3

 # Python3 program to find   # sum of first n terms  def calculateSum(n):      # First term of AP      a1 = 1;             # First term of GP      a2 = 2;             # common ratio of GP      r = 2;             # common difference Of AP      d = 1;      return ((n) * (2 * a1 + (n - 1) * d) /                     2 + a2 * (pow(r, n) - 1) /                    (r - 1));     # Driver Code     # no. of the terms   # for the sum  n = 5;     # Find the Sn  print ("Sum =", int(calculateSum(n)))     # This code is contributed   # by Surendra_Gangwar

## C#

 // C# program to find sum  // of first n terms  using System;     class GFG   {     // Function to calculate the sum  static int calculateSum(int n)  {      // starting number      int a1 = 1, a2 = 2;         // Common Ratio      int r = 2;         // Common difference      int d = 1;         return (n) * (2 * a1 + (n - 1) * d) / 2 + a2 *                (int)(Math.Pow(r, n) - 1) / (r - 1);  }     // Driver code  public static void Main ()   {      // N th term to be find      int n = 5;             // find the Sn      Console.WriteLine("Sum = " + calculateSum(n));  }  }     // This code is contributed   // by inder_verma

## PHP

 

Output:

Sum = 77


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 Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.