На языке паскаль найти произведение 12-ти членов ряда , общий член которого an=1/(3n-2)(3n+1)

Циклик для нахождения(часть программы):
a:=0; p:=1;
for n:=1 to 12 do 
begin
a:=a+1/((3*n-2)*(3*n+1));
p:=p*a;
end;

Вот на С++:

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <cstring>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <climits>typedef unsigned long long ulol;
typedef long double ld;
typedef long long lol;
typedef long int  li;#define mp          make_pair
#define F           first
#define S           second
#define sqr(a)      ( (a) * (a) )
#define pb          push_back
#define INF         999999999
#define ret(a)      cout << endl; system("pause"); return(a)
//#define ret(a)      return(a)using namespace std;int main()
{
    ld x;
    cin >> x;
    x = ( 8 / sqrt( x ) ) + sqrt(x);
    cout << x;
    ret(0);
}

1. домик

screen 12

line (200,120)-(360,320),6,BF

line (240,160)-(320,240),11,BF

line (280,40)-(200,120),2

line (280,40)-(360,120),2

line (200,120)-(360,120),2

paint (280,80),2

 

2. елка

screen 12

line (280,40)-(200,120),2

line (280,40)-(360,120),2

line (200,120)-(360,120),2

paint (280,80),2

line (280,120)-(200,200),2

line (280,120)-(360,200),2

line (200,200)-(360,200),2

paint (280,160),2

line (280,200)-(200,280),2

line (280,200)-(360,280),2

line (200,280)-(360,2800),2

paint (280,240),2

line (240,281)-(320,320),6,BF

line (240,80)-(320,120),8

line (240,160)-(320,200),8

line (240,240)-(320,280),8

circle(260,90),15,1

circle(280,100),15,3

circle(300,110),15,4

circle(260,170),15,5

circle(280,180),15,9

circle(300,190),15,11

circle(260,250),15,12

circle(280,260),15,13

circle(300,270),15,14