Would the following sum give me the following numbers? Would
$\sum_{k=1}^{9}\sum_{j=0}^{9}k*100+10*j+k$
give me the sum of the numbers $101+111+121+131+...+191+202+212+222+...292+303+313+323+...+393...+999$
If not, any advice?
Thanks in advance for the help.
Edit: +k at the end and j starting at $0$,sorry
 A: Yes.  That is what the sum would mean.
$\sum_{k=1}^{9}\sum_{j=0}^{9}k*100+10*j+k = $
$\sum_{k=1}^9([100k + 0 + k]+ [100k + 10 + k] + [100k + 20 + k] + .....+[900k + 90 + k]) =$
$(101 + 111 + 121 + 131 + ...... 191) + $
$(202 + 212 + 222 + 232 + .......292) + $
$......$
$(909+929+929+939+ .....999)$
Alternatively it could also be expressed:
$\sum_{k=1}^9(\sum_{j=0}^9 (k\cdot 100 + 10\cdot j + k)) =$
$\sum_{k=1}^9(\sum_{j=0}^9 k\cdot 100 + \sum_{j=0}^9 10\cdot j + \sum_{j=0}^9 k)=$
$\sum_{k=1}^9 (10\cdot k\cdot 100 + (\sum_{j=0}^9 10\cdot j)+10\cdot k) =$
$\sum_{k=1}^9 (1010k + \sum_{j=0}^9 10\cdot j)$
Can you finish that up to figure out what the sum is?
A: The explicit representation $101+111+\cdots + 999$ corresponds to
\begin{align*}
\sum_{k=1}^{9}\sum_{j=0}^{9}\left(k*100+10*j+k\right)
\end{align*}
This is different to
\begin{align*}
&\sum_{k=1}^{9}\sum_{j=0}^{9}k*100+10*j+k\\
&\qquad=\left(\sum_{k=1}^{9}\sum_{j=0}^{9}k*100\right)+10*j+k\tag{1}\\
&\qquad=\left(10\sum_{k=1}^{9}k*100\right)+10*j+k\\
&\qquad=\left(1000\sum_{k=1}^{9}k\right)+10*j+k\\
&\qquad=1000*(1+2+\cdots+9)+10*j+k\tag{2}
\end{align*}
Note the parentheses in (1) and $j$ and $k$ in the last line (2). This is due to the rules for free and bound variables.
