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:
![S_N = Sum of N Pronic Numbers\\ S_N = [N*\frac{(N - 1)}{2}] + [N*\frac{(N - 1)*(2*N - 1)}{6}]](https://neveropen.co.uk/wp-content/uploads/2023/11/dominic_rubhabha_quicklatex.com-a0848bf16e40c8afc949eea631c1e050_l3.png "Rendered by QuickLaTeX.com")
C++
// C++ implementation to find// sum of first N terms#include <bits/stdc++.h>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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