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

• Difficulty Level : Basic
• Last Updated : 20 Aug, 2022

Given a quadratic series as given below, the task is to find the sum of the 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 the above series, leaving 3 aside, terms starting from 4 to (tn – tn-1) will form an A.P.
Since the formula of the 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 the 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

 

## Javascript

 

Output

Sum = 23

Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.

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