exam question

[saraxmaciel]

How can I do a the sum of: (-1)^(i+1)*[((x-1)^(i))/i]?

I did this :

size_t Sn(size_t &, double &);

int main()
{
size_t n; double x;
cout<<"n: ";
cin>>n;
cout<<"x: ";
cin>>x;
cout<<"S("<<n<<","<<x<<")= " << Sn(n,x)<< endl;
}

size_t Sn(size_t&n, double&x)
{
double sum=1.0;
double a=x-1;
double div=0.0;
double result=0.0;
for(size_t i=1;i<n+1;++i){
result*=a;
div=result/i;
sum+=pow(-1.0,i+1)*div;
return sum;
}

but the only thing that it does is ask for the number n and x

[shadder]

Hi,

Please use code-tags to make your code more readable
http://www.cplusplus.com/articles/jEywvCM9/

size_t Sn(size_t &, double &);

int main()
{
size_t n; double x;
cout<<"n: ";
cin>>n;
cout<<"x: ";
cin>>x;
cout<<"S("<<n<<","<<x<<")= " << Sn(n,x)<< endl;
}

size_t Sn(size_t&n, double&x)
{ 
double sum=1.0;
double a=x-1;
double div=0.0;
double result=0.0;
for(size_t i=1;i<n+1;++i){
result*=a;
div=result/i;
sum+=pow(-1.0,i+1)*div;
return sum;
}

I don't understand what size_t is

[MikeyBoy]

I don't understand what size_t is

http://en.cppreference.com/w/cpp/types/size_t

[shadder]

wow! thanks...

[MikeyBoy]

You're welcome :)