Fortran Factorials

I don't understand at all how you tell the computer to evaluate a complicated factorial expression such as the one given in in the infinite sum of binomial theorem as

Ʃ [n! / k!(n-k)! ] * x^k

where n is the final value of the sum and k is where you are in the loop.

It's supposed to be

INTEGER :: k, n
REAL :: sum, fact, x

ASK INPUT (what are x and n?)

DO k = 0,n
sum = sum + fact*x**k
fact = fact * (n-k)/(k+1)
END DO

Is there a procedure to figure out what the term multiplied by the fact variable should be?
 Recognitions: Science Advisor Look at what the term are summing, for each value of k. When k = 0, it is n! / (0! n!) so k0 = 1 When k = 1, it is n! / (1! (n-1)! so k1 = n = k0 * n / 1 When k = 2, it is n! / (2! (n-2)! so k2 = n(n-1) / 2! = k1 * (n-1) / 2 When k = 3, it is n! / (3! (n-3)! so k3 = n(n-1)(n-2) / 3! = k2 * (n-2) / 3 etc. That is what the program is doing when it updates "fact".
 Hmmm... then fact * (n-k)/(k+1) can't be right because it doesn't match the results of working out all the factorials like that. I checked it in command prompt and it said the sum with x = 2 and n = 2 is 5. This sum represents (1+x)^n which should be 9 in that case. I thought fact * ( (n-k+1) / k ) would work but I'm not getting the right answer with that either.

Fortran Factorials

disregard that. i worked it out on paper and i got 9 but for some reason the program is outputting 5...

If the k0,k1,k2 etc values are the "fact", how is (n-2)/3 = (n-k)/(k+1) for k = 3? That's why I thought it should be ( (n-k+1) / k ).

Recognitions:
 Code:  i=n fact=1 sum=1 DO k= 1,n f = f*i/k*x i = i-1 s = s + f END DO