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

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

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
    
  2. Sum of 1st n natural numbers is given as:
       Sum = (n * (n + 1)) / 2
    

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));
	}
}


Output:

Sum of all the terms in nth row = 132

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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