Thread: series problem stucked in the last step

I am facing some problems sloving this problem...

Code:

sum (2^i * i! )

that means user inputs an number i then the program will calculate 2^i*i! and add the previous i values...

This should be like this

Code:

2^1* 1! + 2^2*2! + 2^3*3! +........+2^n*n!

i am almost solve to the third steps that is calculating ( 2^i * i ! )

but in the last step i cannot add the previous values..i know what to do to get the result that i will need a variable x initialised to 0(zero) and which will increase until it is equal to i .
that is

Code:

for (x=0;x <= i; i++)
{
    x+= (previous value);
}

But i donot know where to put this and i donot know how to make a fuction with this loop..please help me out..i am pasting my existing code that is calculating up to (2^i * i! )

Code:

#include <stdio.h>
#include <conio.h>

int factorial(int i);
int power( int base, int exp );

main()
{
    int n,i,b=2,s=0,multiply;
   	scanf("%d", &i);
	for (n=1;n<=i;n++)
      {
          s=((int) factorial(n));
      }
    multiply = s*power( b, i );
    printf ("The Factorial of %d is\t     : %d\n2 to The Power of %d is\t     : %d\n(Factorial x Exponent) is    : %d\n",i, s,i, power( b, i ), multiply);
    getch();
}

int factorial(int i)
 {
  if (i<=1)
	return(1);
  else
	i=i*factorial(i-1);
	return i;
 }
int power( int base, int exp )
{
	int i, p = 1;
	for( i = 1; i <= exp; i++ )
	     {
             p *= base;
         }
	return p;
}

Code:

int sum = 0;    // cannot hold value more than INT_MAX defined in limits.h 
int i ;
// sum ( 2 ^ i * i! )
for(i = 1; i <= n; i++) {
  sum += power(2,i) * factorial( i );
  // sum = sum + power(2,i) * factorial( i );
}

i,j,k are usually used for counter. Take note that for large value of n, integer may not be big enough.