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, Substract 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) + \sumn + \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++

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// 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;
}

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Java

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// 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));
  
}
}

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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#

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// 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;
 }
}

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PHP

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<?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
?>

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Output:

Sum = 23


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