打印系列 6、28、66、120、190、276 的前 N 个术语……
原文:https://www . geesforgeks . org/print-the-first-n-terms-of-series-6-28-66-120-190-276/
给定一个数字 N ，任务是打印系列 6、28、66、120、190、276 的前 N 个术语，以此类推。 例:
输入:N = 10 T3】输出:6 28 66 120 190 276 378 496 630 780 T6】输入:N = 4 T9】输出: 6 28 66 120
方法:要解决上述问题，我们必须观察以下模式:
通式如下: k *(2 * k–1)，其中，最初 k = 2
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the series
void printSeries(int n)
{
// Initialise the value of k with 2
int k = 2;
// Iterate from 1 to n
for (int i = 0; i < n; i++) {
// Print each number
cout << (k * (2 * k - 1))
<< " ";
// Increment the value of
// K by 2 for next number
k += 2;
}
cout << endl;
}
// Driver Code
int main()
{
// Given number N
int N = 12;
// Function Call
printSeries(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
class GFG{
// Function to print the series
static void printSeries(int n)
{
// Initialise the value of k with 2
int k = 2;
// Iterate from 1 to n
for (int i = 0; i < n; i++)
{
// Print each number
System.out.print(k * (2 * k - 1) + " ");
// Increment the value of
// K by 2 for next number
k += 2;
}
System.out.println();
}
// Driver code
public static void main(String args[])
{
// Given number N
int N = 12;
// Function Call
printSeries(N);
}
}
// This code is contributed by shivaniisnghss2110
Python 3
# Python3 program for the above approach
# Function to print the series
def PrintSeries(n):
# Initialise the value of k with 2
k = 2
# Iterate from 1 to n
for i in range(0, n):
# Print each number
print(k * (2 * k - 1), end = ' ')
# Increment the value of
# K by 2 for next number
k = k + 2
# Driver code
# Given number
n = 12
# Function Call
PrintSeries(n)
# This code is contributed by poulami21ghosh
C
// C# program for the above approach
using System;
class GFG{
// Function to print the series
static void printSeries(int n)
{
// Initialise the value of k with 2
int k = 2;
// Iterate from 1 to n
for(int i = 0; i < n; i++)
{
// Print each number
Console.Write(k * (2 * k - 1) + " ");
// Increment the value of
// K by 2 for next number
k += 2;
}
Console.WriteLine();
}
// Driver code
public static void Main()
{
// Given number N
int N = 12;
// Function call
printSeries(N);
}
}
// This code is contributed by sanjoy_62
java 描述语言
<script>
// javascript program for the above approach
// Function to print the series
function printSeries( n)
{
// Initialise the value of k with 2
let k = 2;
// Iterate from 1 to n
for (let i = 0; i < n; i++) {
// Print each number
document.write((k * (2 * k - 1))
+ " ");
// Increment the value of
// K by 2 for next number
k += 2;
}
document.writeln("<br/>");
}
// Driver Code
// Given number N
let N = 12;
// Function Call
printSeries(N);
// This code is contributed by Rajput-Ji
</script>
Output:
6 28 66 120 190 276 378 496 630 780 946 1128
时间复杂度:O(N) T5】辅助空间: O(1)
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