# Sum of the series 3, 20, 63, 144, ……

Find the sum of first n terms of the given series:

3, 20, 63, 144, .....


Examples:

Input : n = 2
Output : 23

Input : n =4
Output : 230


Approach:
First, we have to find the general term (Tn) of the given series.

series can we written in the following way also:
(3 * 1^2), (5 * 2^2), (7 * 3^2), (9 * 4^2), .......up t n terms
Tn = (General term of series 3, 5, 7, 9 ....) X (General term of series 1^2, 2^2, 3^2, 4^2 ....)
Tn = (3 + (n-1) * 2) X ( n^2 )
Tn = 2*n^3 + n^2


We can write the sum of the series in the following ways:

 Sn = 3 + 20 + 63 + 144 + ........up to the n terms

Sn = 2 * (sum of n terms of n^3 ) + (sum of n terms of n^2)


Following are the formulas of sum of n terms of n^3 and n^2 :

Below is the implementation of the above approach:

## C++

 // C++ program to find the sum of n terms  #include  using namespace std;  int calculateSum(int n)  {      return (2 * pow((n * (n + 1) / 2), 2)) +              ((n * (n + 1) * (2 * n + 1)) / 6);  }  int main()  {      int n = 4;      cout << "Sum = " << calculateSum(n) << endl;      return 0;  }

## Java

 // Java program to find the sum of n terms  import java.io.*;     public class GFG  {      static int calculateSum(int n)      {          return (int)((2 * Math.pow((n * (n + 1) / 2), 2))) +                  ((n * (n + 1) * (2 * n + 1)) / 6);      }             public static void main (String[] args) {                 int n = 4;          System.out.println("Sum = " +  calculateSum(n));             }  }  // This code is contributed by Raj

## Python3

 # Python3 program to find the sum of n terms     def calculateSum(n):      return ((2 * (n * (n + 1) / 2)**2) +             ((n * (n + 1) * (2 * n + 1)) / 6))         #Driver code     n = 4 print("Sum =",calculateSum(n))     # this code is contributed by Shashank_Sharma

## C#

 // C# program to find the sum of n terms  using System;     class GFG  {  static int calculateSum(int n)  {      return (int)((2 * Math.Pow((n * (n + 1) / 2), 2))) +                        ((n * (n + 1) * (2 * n + 1)) / 6);  }     // Driver Code  public static void Main ()   {      int n = 4;      Console.WriteLine("Sum = " + calculateSum(n));  }  }     // This code is contributed by anuj_67

## PHP

 

Output:

Sum = 230


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Improved By : Shashank_Sharma, ash264, R_Raj, vt_m

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