![\displaystyle a_1= \frac{3}{2}; \\ \\ a_{i-1}=\frac{3^{i-1}\cdot (i-1)!}{[2(i-1)]!}; \\ \\ a_i=\frac{3^i\cdot i!}{(2i)!} = \frac{3\cdot3^{i-1}\cdot i\cdot(i-1)!}{2 i\cdot(2i-1)\cdot(2i)!}={ \frac{3i}{2i\cdot(2i-1)}\cdot a_{i-1}=\frac{3}{4i-2}\cdot a_{i-1}](/tpl/images/0829/8194/4eada.png)
1)
#include <iostream>
using namespace std;
int main()
{
int g, k, n=64;
for (g=0; g<=n; g++)
{
for (k=0; k<=n/2; k++)
{
if (2*g + 4*k == n)
{
printf ("Gycu - %d. \nKpoluk - %d.\n\n", g, k);
}
}
}
return 0;
}
2)#include <iostream>
using namespace std;
long double fact(int N)
{
if(N < 0)
return N;
if (N == 0)
return 1;
else
return N * fact(N - 1);
}
int main()
{
int N, M;
setlocale(0,"");
cout << "N=";
cin >> N;
cout<<"M=";
cin>>M;
cout<<"N!-M!=" << fact(N)-fact(M) << endl << endl;
return 0;
}
begin
readln(c);
for b:=1 to c do
begin
readln(r);
If r mod 4 =0 then v:=v+1;
end;
writeln(v);
end.