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Calculate the next sum recursive

thedoctor2707 (5)
Hello I have to calculate the next sum recursively :

S= 1! + 1/2! + 3! + 1/4! + 5! + 1/6! + ......+ 1/(2n)! + (2n+1)!
So I split that into 2 sums:
S1 = 1! + 3! + 5! + ..... + (2n+1)! ->> S(n - 1) + n!, if n is odd

S2 = 1/2! + 1/4! + 1/6! + ...... + 1/(2n)! ->> S(n - 1) + 1 / (n!), if n is even

And I have to choose the number, even or odd.
This is what I got so far :


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 #include <iostream>
using namespace std;

int factorial (int n)
{
	if (n==0) 
		return 1;
	else 
		return n*factorial(n-1);    //function for the factorial
}

float SumaPar (int n)
{
	if (n==0)
		return 0;
	else
		return SumaPar(n-1) + (1.0/factorial(n));   //function for the even numbers
}

float SumaImpar (int n)
{
	if (n==0)
         return 0;
	else
         return SumaImpar(n-1) + (factorial(n));  //function for the odd numbers
}

int main ()

{
int n;
float S=0;

cout<<"Introduceti n=";
cin>>n;

    {
    	if (n%2==0) 
    	{
    		S= (SumaImpar(n) + SumaPar(n)) - 1;
		}
		else
		{
		
		    S= (SumaImpar(n) + SumaPar(n-1)) - 1 ;
		}
	}
	
	cout<<"Suma este : "<<S;
	
        system("pause");
}


All works fine, but....instead of my sum it gives me this :

S=1!+2!+3!+4!+......+n! + 1/2! + 1/3! + 1/4! + 1/5! + ..... + 1/(2n+1)!

Any help will be apreciated.
thedoctor2707 (5)
No it's the wrong sum, example :
n=7 --> 1! + 1/2! + 3! + 3/4! + 5! + 5/6! + 7! + 7/8! = 5167.63.
Does this look like :

n=7--> 1! + 1/2! + 3! + 1/4! + 5! + 1/6! + 7! ???
Without for, and while - already done that.

Sry. new to this forum.The code is written in romanian.
Last edited on
thedoctor2707 (5)
Does this look the same?

1! + 1/2! + 3! + 3/4! + 5! + 5/6! + 7! + 7/8! = 5167.63.

1!+ 1/2!+ 3!+ 1/4!+ 5!+ 1/6!+ 7!=5167.54

Don't know what math you've been doing but, 3!+1/4! does not equal 3/4!
Last edited on
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