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

Question

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?

Answer 1

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

$$\sum_{i=1}^n (10^{i+1}-10^i)=10^{n+1}-10$$

Answer 2

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.

Answer 3

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

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

Answer 4

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