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 sumsum= (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 * n^{2}+ 1)

The proof for the formula is given below:

**Prerequisite:**

- 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** - 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 thenthrow bep. Herep= 2 + 4 + 6 + .....nterms For the givenAPseries,a= 2,d= 2. Using the above formula for the sum ofnterms 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)throw beq. Hereq= 2 + 4 + 6 + .....n-1terms For the givenAPseries,a= 2,d= 2. Using the above formula for the sum ofn-1terms of the AP series, we get,q = n * (n - 1)Now, Sum of all the terms in the nth row = sum of 1stpnatural numbers - sum of 1stqnatural numbers =(p * (p + 1)) / 2 - (q * (q + 1)) / 2Substituting the values ofpandqand then solving the equation, we will get,Sum of all the terms in the nth row = n * (2 * n^{2}+ 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

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