Program to print series 0 2 6 12 20 30 42 ...N

 Problem statement:- Program to print series 0 2 6 12 20 30 42 ...N.

 Data requirement:-

   Input Data:- n

  Output Data:-i

  Additional Data:- Nothing.

Program in C
  
Here is the source code of the C Program to print series 0 2 6 12 20 30 42 ...N.

Code:

#include<stdio.h>
int main()
{
    int n,i;
    printf("Enter the range of number(Limit):");
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        printf("%d ",(i*i)-i);
    }
}

Input/Output:
Enter the range of number(Limit):7
0 2 6 12 20 30 42

Program in C++

Here is the source code of the C++ Program to print series 0 2 6 12 20 30 42 ...N.

Code:

#include<iostream>
using namespace std;
int main()
{
    int n,i;
    cout<<"Enter the range of number(Limit):";
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cout<<(i*i)-i<<" ";
    }
}

Input/Output:
Enter the range of number(Limit):5
0 2 6 12 20

Program in Java

Here is the source code of the Java Program to print series 0 2 6 12 20 30 42 ...N.

Code:

import java.util.Scanner;
public class Print_Series11 {

public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n,i;
   System.out.printf("Enter the range of number(Limit):");
    n=sc.nextInt();
    for(i=1;i<=n;i++)
    {
    System.out.print((i*i)-i+" ");
    }
    sc.close();
}
}

Input/Output:
Enter the range of number(Limit):6
0 2 6 12 20 30 

Program in Python

Here is the source code of the Python Program to print series 0 2 6 12 20 30 42 ...N.

Code:

n=int(input("Enter the range of number(Limit):"))
i=1
while i<=n:
    print((i*i)-i,end=" ")
    i+=1

Input/Output:
Enter the range of number(Limit):7
0 2 6 12 20 30 42 


Post a Comment

1 Comments

Please do not Enter any spam link in the comment box