# Sum of first N terms of Quadratic Sequence 3 + 7 + 13 + …

Given a quadratic series as given below, the task is to find sum of first n terms of this series.

Sn = 3 + 7 + 13 + 21 + 31 + ….. + upto n terms

Examples:

Input: N = 3
Output: 23

Input: N = 4
Output: 44


Approach:
Let the series be represented as

Sn = 3 + 7 + 13 + ... + tn

where

• Sn represents the sum of the series till n terms.
• tn represents the nth term of the series.

Now, to formulate the series, the elements need to be formed by taking the difference of the consecutive elements of the series.

Equation 1: Sn = 3 + 7 + 13 + 21 + 31 +…..+ tn-1 + tn
Equation 2: Sn = 0 + 3 + 7 + 13 + 21 + 31 + …… + tn-1 + tn
(writing the above series by shifting all elements to right by 1 position)

Now, Subtract Equation 2 from Equation 1 i.e. (Equation 1 – Equation 2)

Sn – Sn = (3 – 0) + (7 – 3) + (13 – 7) + (31 – 21) + …… + (tn- tn-1) – tn
=> 0 = 3 + 4 + 6 + 8 + 10 + …… + (tn – tn-1) – tn

In above series, leaving 3 aside, terms starting from 4 to (tn – tn-1) will form an A.P.

Since formula of sum of n terms of A.P. is:

Sn = n*(2*a + (n – 1)*d)/2

which implies,

In series: 4 + 6 + 8 + … + (tn – tn-1)
AP is formed with (n-1) terms.

Hence,

Sum of this series: (n-1)*(2*4 + (n-2)*2)/2

Therefore, the original series:
0 = 3 + (n-1)*(2*4 + (n-2)*2)/2 – tn
where tn = n^2 + n + 1 which is the nth term.

Therefore,

Sum of first n terms of series will be:

tn = n^2 + n + 1
Sn = (n^2) + n + (1)
Sn = n*(n+1)*(n+2)/6 + n*(n+1)/2 + n
Sn = n*(n^2 + 3*n + 5)/3

Below is the implementation of above approach:

## C++

 // C++ program to find sum of first n terms     #include  using namespace std;     int calculateSum(int n)  {      // Sum = n*(n^2 + 3*n + 5)/3      return n * (pow(n, 2) + 3 * n + 5) / 3;  }     int main()  {      // number of terms to be included in the sum      int n = 3;         // find the Sum      cout << "Sum = " << calculateSum(n);         return 0;  }

## Java

 // Java program to find sum of first n terms  import java.util.*;     class solution  {  //function to calculate sum of n terms of the series  static int calculateSum(int n)  {      // Sum = n*(n^2 + 3*n + 5)/3      return n * (int)  (Math.pow(n, 2) + 3 * n + 5 )/ 3;  }     public static void main(String arr[])  {      // number of terms to be included in the sum      int n = 3;         // find the Sum      System.out.println("Sum = " +calculateSum(n));     }  }

## Python3

 # Python 3 program to find sum   # of first n terms  from math import pow    def calculateSum(n):             # Sum = n*(n^2 + 3*n + 5)/3      return n * (pow(n, 2) + 3 * n + 5) / 3    if __name__ == '__main__':             # number of terms to be included      # in the sum      n = 3        # find the Sum      print("Sum =", int(calculateSum(n)))     # This code is contributed by  # Sanjit_Prasad

## C#

 // C# program to find sum of first n terms  using System;  class gfg  {   public double calculateSum(int n)   {      // Sum = n*(n^2 + 3*n + 5)/3      return (n * (Math.Pow(n, 2) + 3 * n + 5) / 3);    }  }     //driver code  class geek  {   public static int Main()    {       gfg g = new gfg();      // number of terms to be included in the sum      int n = 3;      //find the Sum      Console.WriteLine( "Sum = {0}", g.calculateSum(n));      return 0;   }  }

## PHP

 

Output:

Sum = 23


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