# 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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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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