Program C++ untuk mencari jumlah dari deret 1-1/2+1/3-1/4+...+1/n

Algoritma :

Deklarasi

n                      : integer

rumus,jumlah,total   : float

Deskripsi

jumlah = 0;

total = 0;

rumus = -1;

for j <= 1 to n do

rumus =(rumus*(-1));

total = rumus/j;

jumlah += total;

if (j==1)

              write(total)

if (j>1)

then

              write(+ total)

endif

then

write (jumlah)

endfor

Berikut programnya :

#include <iostream.h>

class hitung

     {

     public:

      int proses();

      void input();

     private:

      int n;

      float rumus,jumlah,total;

     };

     void hitung::input()

     {

      cin>>n;

      cout<<endl;

     }

     int hitung::proses()

     {

      jumlah=0;

      total=0;

      rumus=-1;

      for(int j=1; j<=n; j++)

      {

      rumus=(rumus*(-1));

      total=rumus/j;

      jumlah+=total;

      if(j==1)

      cout<<"("<<total<<")";

      if(j>1)

      cout<<"+("<<total<<")";

      }

     cout<<endl<<endl<<"hasil penjumlahan deret = "<<jumlah;

     return jumlah;

     }

     int main()

     {

     cout<<"program sederhana menghitung jumlah dari rumus 1-(1/2)+(1/3)-(1/4)+...+(1/n)"<<endl<<endl;

     cout<<"tentukan nilai n : ";

     hitung deret;

     deret.input();

     deret.proses();

      

     return 0;

}

Explanation:
The program above is used to calculate the number of runs yatiu 1-1 / 2 + 1 / 3 - 1 / 4 +...+ 1 / n by using a recursive function which will call itself the return amount;. In this program using for loop and if the election. The process of the program will execute the first value of j = 1 as initial value. After that will run the formula specified in the program and then conduct the election by using if that is if (j == 1) then the total will appear and if (j> 1) will add the total. Order number appears the program will call itself the return number; such as the above program.