Given summation is an AGP.
Let $\mathcal{S}$ denote the given summation.
$\mathcal{S}=3+2.3^2+3.3^3...+10.3^{10}$ $3 \mathcal{S}=~~~~~~~~3^2+2.3^3+...+9.3^{10}+10.3^{11}$ Subtracting these, we get: $-2\mathcal{S}=\color{#D61F06}{3+3^2+...+3^{10} }-3^{11}$ $\color{#D61F06}{Red}$ terms are sum of a GP. Calculating it and few manipulations gives: $\mathcal{S}=\dfrac{19.3^{11}+3}{4}$ $\Large \Rightarrow \mathcal{S}=\boxed{841449}$

Simple Cpp works,

include<iostream>

include<math.h>

using namespace std; int main() { int fact=1,i,num;

     float sum=0 ,term=1;


      cout<<"Enter the number of terms";

      cin>>num;

       for(i=1;i<=num;i++)
{
    term=i*pow(3,i);
    sum=sum+term;
}
cout<<"sum="<<sum;
fflush(stdin);
getchar();
return 0;

}