Hi, new to the forum but I have read and understand u guys aren't here to do this for me and I know that and want to learn how to do this.

My goal is to write a program that to compute the following sum: 1+2*3+3*4*5+4*5*6*7+5*6*7*8*9+6*7*8*9*10*11+...+n* (n+1)*(n+2)*...*(2n-1)

Now me and some people figured this to be the same thing as

(2n-1)!/(n-1)!

here is what I have come up with so far but I am sure it's not correct any tips, suggestions, or constructive criticism is welcomed.

I know this isnt right b/c if n = 3 the answer output from the program should be 67 b/cCode:`#include <iostream>`

using namespace std;

int main()

{

int k, n, total=0, sum=0;

cout << "Enter a whole number: \n";

cin >> n;

k = n;

while (n>=1)

{

while (k<=2*n-1)

{

sum = k * ++k;

}

total= total+sum;

k = --n;

--n;

}

cout << "The answer to the equation for your number is ";

cout << total << endl;

return 0;

}

(5!/2!) + (3!/1!) + (1!/0!) = 67 but I get 72 when I run the program.

Thanks and sorry for the long post, me and my friends are just noobs trying to learn from mistakes.