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Find the sum of all the terms in the n-th row of the given series

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Find the sum of all the terms in the nth row of the series given below. 
 

               1  2
            3  4  5  6
         7  8  9 10 11 12
     13 14 15 16 17 18 19 20
    ..........................
   ............................
             (so on)

Examples: 
 

Input : n = 2
Output : 18
terms in 2nd row and their sum
sum = (3 + 4 + 5 + 6) = 18

Input : n = 4
Output : 132

 

Recommended Practice

Naive Approach: Using two loops. Outer loop executes for i = 1 to n times. Inner loop executes for j = 1 to 2 * i times. Counter variable k to keep track of the current term in the series. When i = n, the values of k are accumulated to the sum. 
Time Complexity: O(k), where k is the total number of terms from the beginning till the end of the nth row.
Efficient Approach: The sum of all the terms in the nth row can be obtained by the formula: 
 

 Sum(n) = n * (2 * n2 + 1)

The proof for the formula is given below:
Prerequisite:
 

  1. Sum of n terms of an Arithmetic Progression series with a as the first term and d as the common difference is given as: 
     
   Sum = (n * [2*a + (n-1)*d]) / 2
  1.  
  2. Sum of 1st n natural numbers is given as: 
     
   Sum = (n * (n + 1)) / 2
  1.  

Proof: 
 

Let the number of terms from the beginning 
till the end of the nth row be p.
Here p = 2 + 4 + 6 + .....n terms
For the given AP series, a = 2, d = 2.
Using the above formula for the sum of
n terms of the AP series, we get,

     p = n * (n + 1)

Similarly, let the number of terms from the 
beginning till the end of the (n-1)th row be q.
Here q = 2 + 4 + 6 + .....n-1 terms
For the given AP series, a = 2, d = 2.
Using the above formula for the sum of
n-1 terms of the AP series, we get,

     q = n * (n - 1)

Now,
Sum of all the terms in the nth row 
           = sum of 1st p natural numbers - 
             sum of 1st q natural numbers
    
           = (p * (p + 1)) / 2 - (q * (q + 1)) / 2

Substituting the values of p and q and then solving
the equation, we will get,

Sum of all the terms in the nth row = n * (2 * n2 + 1)

 

C++




// C++ implementation to find the sum of all the
// terms in the nth row of the given series
#include <bits/stdc++.h>
 
using namespace std;
 
// function to find the required sum
int sumOfTermsInNthRow(int n)
{
    // sum = n * (2 * n^2 + 1)
    int sum = n * (2 * pow(n, 2) + 1);
    return sum;
}
 
// Driver program to test above
int main()
{
    int n = 4;
    cout << "Sum of all the terms in nth row = "
         << sumOfTermsInNthRow(n);
    return 0;
}


Java




// Java implementation to find the sum of all the
// terms in the nth row of the given series
 
import static java.lang.Math.pow;
 
class Test {
    // method to find the required sum
    static int sumOfTermsInNthRow(int n)
    {
        // sum = n * (2 * n^2 + 1)
        int sum = (int)(n * (2 * pow(n, 2) + 1));
        return sum;
    }
 
    // Driver method
    public static void main(String args[])
    {
        int n = 4;
        System.out.println("Sum of all the terms in nth row = "
                           + sumOfTermsInNthRow(n));
    }
}


Python3




# Python 3 implementation to find
# the sum of all the terms in the
# nth row of the given series
from math import pow
 
# function to find the required sum
def sumOfTermsInNthRow(n):
     
    # sum = n * (2 * n^2 + 1)
    sum = n * (2 * pow(n, 2) + 1)
    return sum
 
# Driver Code
if __name__ == '__main__':
    n = 4
    print("Sum of all the terms in nth row =",
                   int(sumOfTermsInNthRow(n)))
 
# This code is contributed
# by Surendra_Gangwar


C#




// C# implementation to find the sum of all the
// terms in the nth row of the given series
using System;
 
class Test {
    // method to find the required sum
    static int sumOfTermsInNthRow(int n)
    {
        // sum = n * (2 * n^2 + 1)
        int sum = (int)(n * (2 * Math.Pow(n, 2) + 1));
        return sum;
    }
 
    // Driver method
    public static void Main()
    {
        int n = 4;
        Console.Write("Sum of all the terms in nth row = "
                      + sumOfTermsInNthRow(n));
    }
}
 
// This code is contributed by vt_m.


PHP




<?php
// PHP implementation to find
// the sum of all the terms in
// the nth row of the given series
 
// function to find the required sum
function sumOfTermsInNthRow($n)
{
     
    // sum = n * (2 * n^2 + 1)
    $sum = $n * (2 * pow($n, 2) + 1);
    return $sum;
}
 
    // Driver Code
    $n = 4;
    echo "Sum of all the terms in nth row = ",
                        sumOfTermsInNthRow($n);
 
// This code is contributed by ajit
?>


Javascript




<script>
// javascript implementation to find the sum of all the
// terms in the nth row of the given series
 
// function to find the required sum
function sumOfTermsInNthRow( n)
{
 
    // sum = n * (2 * n^2 + 1)
    let sum = n * (2 * Math.pow(n, 2) + 1);
    return sum;
}
 
// Driver program to test above
 
    let n = 4;
    document.write( "Sum of all the terms in nth row = "
         + sumOfTermsInNthRow(n));
          
// This code is contributed by aashish1995
 
</script>


Output: 
 

Sum of all the terms in nth row = 132

Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

 



Last Updated : 31 May, 2022
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