级数之和 1 / 1 + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + – – – – – – 最多 n 项。
给定一个正整数 n,求级数的总和,直到 n 项。
例子 :
Input : n = 4
Output : 3.91667
Series : (1 / 1) + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + (1 + 2 + 3 + 4) / (1 * 2 * 3 * 4)
= 1 / 1 + 3 / 2 + 6 / 6 + 10 / 24
= 3.91667
Input : n = 6
Output : 4.07083
Series : (1 / 1) + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + (1 + 2 + 3 + 4) / (1 * 2 * 3 * 4) + (1 + 2 + 3 + 4 + 5) / (1 * 2 * 3 * 4 * 5) + (1 + 2 + 3 + 4 + 5 + 6) / (1 * 2 * 3 * 4 * 5 * 6)
= 1 / 1 + 3 / 2 + 6 / 6 + 10 / 24 + 15 / 120 + 21 / 720
= 4.07083
C++
// CPP program to find sum of series.
#include
using namespace std;
double sumOfSeries(int n)
{
double res = 0.0 ;
int sum = 0, prod = 1;
for (int i = 1 ; i <= n ; i++)
{
sum += i;
prod *= i;
res += ((double)sum / prod);
}
return res;
}
// Driver Code
int main()
{
int n = 4 ;
cout << sumOfSeries(n) ;
return 0;
}
Java
// Java program to find sum of series.
class GFG
{
static double sumOfSeries(int n)
{
double res = 0.0;
int sum = 0, prod = 1;
for (int i = 1; i <= n; i++) {
sum += i;
prod *= i;
res += ((double)sum / prod);
}
return res;
}
// Driver code
public static void main(String arg[]) {
int n = 4;
System.out.println(sumOfSeries(n));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python program to
# find sum of series.
def sumOfSeries(n) :
res = 0.0
sum = 0
prod = 1
for i in range(1, n + 1) :
sum = sum + i
prod = prod * i
res = res + (sum / prod)
return res
# Driver Code
n = 4
print (round(sumOfSeries(n), 5))
# This code is contributed by
# Manish Shaw(manishshaw1)
C#
// C# program to find sum of series.
using System;
class GFG {
static double sumOfSeries(int n)
{
double res = 0.0;
int sum = 0, prod = 1;
for (int i = 1; i <= n; i++) {
sum += i;
prod *= i;
res += ((double)sum / prod);
}
return res;
}
// Driver code
public static void Main()
{
int n = 4;
Console.Write(sumOfSeries(n));
}
}
// This code is contributed by vt_m.
PHP
Javascript
输出 :
3.91667