给定一个数N的任务是找到以下系列总和高达第n项。
3, 7, 13, 21, 31, ….
例子:
Input : N = 3
Output : 23
Input : N = 25
Output : 5875
方法:
减去以上两个等式,我们得到:
下面是上述方法的实现:
C++
// C++ Program to find the sum of given series
#include
#include
using namespace std;
// Function to calculate sum
int findSum(int n)
{
// Return sum
return (n * (pow(n, 2) + 3 * n + 5)) / 3;
}
// Driver code
int main()
{
int n = 25;
cout << findSum(n);
return 0;
}
Java
// Java program to find sum of
// n terms of the given series
import java.util.*;
class GFG
{
static int calculateSum(int n)
{
// returning the final sum
return (n * ((int)Math.pow(n, 2) + 3 *
n + 5)) / 3;
}
// Driver Code
public static void main(String arr[])
{
// number of terms to
// find the sum
int n = 25;
System.out.println(calculateSum(n));
}
}
// This code is contributed
// by Surendra_Gangwar
Python 3
# Python program to find the
# sum of given series
# Function to calculate sum
def findSum(n):
# Return sum
return (n*(pow(n, 2)+3 * n + 5))/3
# driver code
n = 25
print(int(findSum(n)))
C#
// C# program to find
// sum of n terms of
// the given series
using System;
class GFG
{
static int calculateSum(int n)
{
// returning the final sum
return (n * ((int)Math.Pow(n, 2) + 3 *
n + 5)) / 3;
}
// Driver Code
public static void Main()
{
// number of terms to
// find the sum
int n = 25;
Console.WriteLine(calculateSum(n));
}
}
// This code is contributed
// by inder_verma.
PHP
Javascript
输出:
5875
时间复杂度: O(1)