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

  • Difficulty Level : Basic
  • Last Updated : 08 Apr, 2021

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

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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 = \sum   (n^2) + \sum   n + \sum   (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 <bits/stdc++.h>
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




<?php
// PHP program to find sum
// of first n terms
 
function calculateSum($n)
{
    // Sum = n*(n^2 + 3*n + 5)/3
    return $n * (pow($n, 2) + 3 *
                     $n + 5) / 3;
}
 
// Driver Code
 
// number of terms to be
// included in the sum
$n = 3;
 
// find the Sum
echo "Sum = " . calculateSum($n);
 
// This code is contributed by mits
?>

Javascript




<script>
 
// Javascript program to find sum of first n terms
 
// Function to find the quadratic
// equation whose roots are a and b
function calculateSum(n)
{
     
    // Sum = n*(n^2 + 3*n + 5)/3
    return n * (Math.pow(n, 2) + 3 * n + 5 ) / 3;
}
 
// Driver Code
 
// Number of terms to be
// included in the sum
var n = 3;
 
// Find the Sum
 
document.write("Sum = " + calculateSum(n));
 
// This code is contributed by Ankita saini
     
</script>
Output: 
Sum = 23

 




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