# Sum of the first N Pronic Numbers

• Last Updated : 01 Jul, 2020

Given a number N, the task is to find the sum of the first N Pronic Numbers.

The numbers that can be arranged to form a rectangle are called Rectangular Numbers (also known as Pronic numbers). The first few rectangular numbers are:
0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462 . . . .

Examples:

Input: N = 4
Output: 20
Explanation:
0, 2, 6, 12 are the first 4 pronic numbers.

Input: N = 3
Output: 8

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Let, the Nth term be denoted by TN. This problem can easily be solved by splitting each term as follows:     Therefore:

SN = Sum of N Pronic Numbers Below is the implementation of the above approach:

## C++

 // C++ implementation to find // sum of first N terms#include using namespace std;  // Function to calculate the sumint calculateSum(int N){      return N * (N - 1) / 2           + N * (N - 1)                 * (2 * N - 1) / 6;}  // Driver codeint main(){    int N = 3;      cout << calculateSum(N);      return 0;}

## Java

 // Java implementation implementation to find // sum of first N terms class GFG{       // Function to calculate the sum static int calculateSum(int N) {           return N * (N - 1) / 2 + N * (N - 1) *           (2 * N - 1) / 6; }       // Driver code public static void main (String[] args) {     int N = 3;           System.out.println(calculateSum(N)); } }   // This code is contributed by Pratima Pandey

## Python3

 # Python3 implementation to find # sum of first N terms  # Function to calculate the sumdef calculateSum(N):      return (N * (N - 1) // 2 +             N * (N - 1) * (2 *                  N - 1) // 6);  # Driver codeN = 3;print(calculateSum(N));  # This code is contributed by Code_Mech

## C#

 // C# implementation implementation to find // sum of first N terms using System;class GFG{       // Function to calculate the sum static int calculateSum(int N) {           return N * (N - 1) / 2 + N * (N - 1) *                        (2 * N - 1) / 6; }       // Driver code public static void Main() {     int N = 3;           Console.Write(calculateSum(N)); } }   // This code is contributed by Code_Mech

Output:

8


Time complexity: O(1).

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