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! 
 A: Without derivatives: Notice that the sum is a sum of geometric sums, namely
$$
2+2\cdot 2^2  + 3\cdot 2^3+ \dots + n2^n = 
(2+ \dots + 2^n) + (2^2 + \dots + 2^n) + (2^3 + \dots + 2^n) + \dots + (2^n)
$$
i.e. we use the $2^k$ term $k$ times.
A: Note:
$$S_n=1\cdot 2^1+2\cdot 2^2+3\cdot 2^3+\cdots +n\cdot 2^n\\
2S_n=1\cdot 2^2+2\cdot 2^3+3\cdot 2^4+\cdots +(n-1)\cdot 2^n+n\cdot 2^{n+1}\\
2S_n-S_n=-1\cdot 2^1-1\cdot 2^2-1\cdot 2^3-\cdots -1\cdot 2^n+n\cdot 2^{n+1} \Rightarrow\\
S_n=-(2+2^2+2^3+\cdots+2^n)+n\cdot 2^{n+1}=2+2^{n+10} \Rightarrow \\
-\frac{2(2^n-1)}{2-1}+n\cdot 2^{n+1}=2+2^{n+10} \Rightarrow \\
-2\cdot 2^n+2+n\cdot 2\cdot 2^n=2+2^{10}\cdot 2^n \Rightarrow \\
n\cdot 2\cdot 2^n=2^{10}\cdot 2^n+2\cdot 2^n \Rightarrow \\
2n=2^{10}+2 \Rightarrow \\
n=2^9+1=513 \Rightarrow \\
5+1+3=9.$$
A: Hint
$$S\triangleq\sum_{r=1}^n rx^{r}=x\sum_{r=1}^n rx^{r-1}$$where $$\sum_{r=1}^n rx^{r-1}={d\over dx}\sum_{r=1}^n x^{r}$$
An alternative way
$$S{=x+\sum_{r=2}^{n}rx^r\\=x+\sum_{r=1}^{n-1}(r+1)x^{r+1}\\=x-(n+1)x^{n+1}+\sum_{r=1}^{n}(r+1)x^{r+1}
\\=x-(n+1)x^{n+1}+x\sum_{r=1}^{n}(r+1)x^{r}
\\=x-(n+1)x^{n+1}+x\sum_{r=1}^{n}rx^{r}
+x\sum_{r=1}^{n}x^{r}
\\=x-(n+1)x^{n+1}+xS+x^2{1-x^n\over 1-x}
}$$
A: There is this sum technique you can use.
\begin{align*} S_n+(n+1)2^{n+1} & =  S_{n+1} =  \sum_{k=1}^{n+1} k2^k \\ & =  2+\sum_{k=2}^{n+1} k2^k = 2+\sum_{k=1}^n (k+1)2^{k+1} \\ & = 2+2S_n+4(2^n-1) \end{align*} 
Now, just take terms with $S_n$ on one side, and other terms on the other side.
