The sum of all digits of n for which $\sum^n_{r=1} r. 2^r=2+2^{n+10}$ If we start substituting values, we get 
$$S_n= 2+8+24+64......n2^n$$
The nth term of the series will be 
$$T_n = (n)(2^n)$$
So the sum would be $$\sum T_n=\sum n 2^n$$
Cause that’s how usually solve it. The exponent is throwing me off, and I don’t know what to do. 
Thanks! 

Note:
$$S_n=1\cdot 2^1+2\cdot 2^2+3\cdot 2^3+\cdots +n\cdot 2^n\\
2S_n=1\cdot 2^2+2\cdot 2^3+3\cdot 2^4+\cdots +(n-1)\cdot 2^n+n\cdot 2^{n+1}\\
2S_n-S_n=-1\cdot 2^1-1\cdot 2^2-1\cdot 2^3-\cdots -1\cdot 2^n+n\cdot 2^{n+1} \Rightarrow\\
S_n=-(2+2^2+2^3+\cdots+2^n)+n\cdot 2^{n+1}=2+2^{n+10} \Rightarrow \\
-\frac{2(2^n-1)}{2-1}+n\cdot 2^{n+1}=2+2^{n+10} \Rightarrow \\
-2\cdot 2^n+2+n\cdot 2\cdot 2^n=2+2^{10}\cdot 2^n \Rightarrow \\
n\cdot 2\cdot 2^n=2^{10}\cdot 2^n+2\cdot 2^n \Rightarrow \\
2n=2^{10}+2 \Rightarrow \\
n=2^9+1=513 \Rightarrow \\
5+1+3=9.$$
