Evaluate $\sum^{n}_{i=1}(10^{i+1}-10^i)$ 
Evaluate $$\sum^{n}_{i=1}(10^{i+1}-10^i)$$

Here's what I did
$$\sum^{n}_{i=1}(10^{i+1}-10^i)
\\ = 10(\sum^n_{i=1} 1^{i+1}-1^i)
\\ = -10(\sum^n_{i=1} 1^i-1^{i+1})
\\= -10(1^n-1^{n+1})
\\= 10^{n+1}-10^n$$
Is this correct?
 A: There are multiple conceptual errors in the OP.
The sum of interest is of the general form
$$\begin{align}
\sum_{i=1}^n (a_{i+1}-a_i)&=(a_{2}-a_1)+(a_3-a_2)+(a_4-a_3)+\cdots +(a_n-a_{n-1}) +(a_{n+1}-a_n)\\\\&=a_{n+1}-a_1
\end{align}$$
For $a_i=10^i$, we obtain 
$$\sum_{i=1}^n (10^{i+1}-10^i)=10^{n+1}-10$$
A: The wrong step you made is factoring $10$ out of the summation. You basically said this:
$$10^{i+1}-10^i=10(1^{i+1}-1^i)$$
Substitute $i=1$ into the above equation. Do you see why it doesn't work know?
What you were supposed to notice is that you can do this by telescoping, as @OliverOloa pointed out. For more on telescoping series, look at this example from Wikipedia.
A: It is not correct.
One may observe that we are dealing with a telescoping sum:
$$
\sum_{i=1}^n\left(u_{i+1}-u_i \right)=u_{n+1}-u_1,
$$ here we have $u_i=10^i$ giving

$$
\sum^{n}_{i=1}(10^{i+1}-10^i)=10^{n+1}-10^1.
$$

A: It should be as follows,you have made a mistake.
$$\sum _{ i=1 }^{ n } (10^{ i+1 }-10^{ i })=\sum _{ i=1 }^{ n }{ { 10 }^{ i }\left( 10-1 \right) =9\sum _{ i=1 }^{ n }{ { 10 }^{ i } }  } =9\left( 10+{ 10 }^{ 2 }+...+{ 10 }^{ n } \right) =9\frac { 10\left( 1-{ 10 }^{ n } \right)  }{ 1-10 } ={ 10 }^{ n+1 }-10$$
