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

     $$    Sn = \sum_{n=1}^{n} T_{n} $$

      $$      Sn = 2 \times \sum_{n=1}^{n} n^{3} + \sum_{n=1}^{n} n^{2}  $$

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 :

      $$       \sum_{n=1}^{n} n^{3} = \left[\frac{n \times \big(n + 1 \big) }{2} \right]^{2}          $$  $$       \sum_{n=1}^{n} n^{2} = \frac{n \times \big(n + 1 \big) \times \big(2*n + 1 \big) }{6}           $$

 Total = 2 \times \left[\frac{n \times \big(n + 1 \big) }{2} \right]^{2}  + \frac{n \times \big(n + 1 \big) \times \big(2*n + 1 \big) }{6}

Below is the implementation of the above approach:

C++

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// C++ program to find the sum of n terms
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?php
// PHP program to find the 
// sum of n terms
  
function calculateSum($n)
{
    return (2 * pow(($n * ($n + 1) / 2), 2)) + 
           (($n * ($n + 1) * (2 * $n + 1)) / 6);
}
  
// Driver Code
$n = 4;
echo "Sum = " , calculateSum($n);
  
// This code is contributed by ash264
?>

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Output:

Sum = 230


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Improved By : Shashank_Sharma, ash264, R_Raj, vt_m