# Sum of the first N Pronic Numbers

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

Approach:

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

Therefore:

## 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 termsclass GFG{     // Function to calculate the sumstatic int calculateSum(int N){         return N * (N - 1) / 2 + N * (N - 1) *        (2 * N - 1) / 6;}     // Driver codepublic 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 termsusing System;class GFG{     // Function to calculate the sumstatic int calculateSum(int N){         return N * (N - 1) / 2 + N * (N - 1) *                        (2 * N - 1) / 6;}     // Driver codepublic static void Main(){    int N = 3;         Console.Write(calculateSum(N));}} // This code is contributed by Code_Mech

## Javascript

 // JS implementation to find// sum of first N terms  // Function to calculate the sumfunction calculateSum(N){     return  Math.floor(N * (N - 1) / 2) + Math.floor(N * (N - 1) * (2 * N - 1) / 6);} // Driver codelet N = 3;console.log(calculateSum(N));  // This code is contributed by phasing17

Output
8

Time complexity: O(1).
Auxiliary Space: O(1)

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