1. ## Algebra

I am doing proof by induction for my discrete math class by I'm horrible at algebra and I can't figure this last part out. Thanks in advance for the advice.

I need to make: (10^k) -1 + (9)(10^k)

look like this: 10^(k+1) -1

I also need help with this one:

I need to make: 1 - 1/(k+1) + 1/(k+1)(k+2) (1 is numerator, rest is Denom)

look like this: 1 - 1/(k+2)

2. Originally Posted by jzellt
...

I need to make: (10^k) -1 + (9)(10^k)

look like this: 10^(k+1) -1

...
$10^k - 1 +9 \cdot 10^k = 10^k - 1 +(10 -1) \cdot 10^k = 10^k - 1 +10 \cdot 10^k - 10^k= -1 + 10 \cdot 10^k = 10^{k+1}-1$

3. Thank you! Any advice for the second problem?

4. Originally Posted by jzellt
...

I also need help with this one:

I need to make: 1 - 1/(k+1) + 1/(k+1)(k+2) (1 is numerator, rest is Denom)

look like this: 1 - 1/(k+2)
$1-\frac1{k+1}+\frac1{(k+1)(k+2)}= 1-\frac{k+2}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)} = 1- \frac{k+2-1}{(k+1)(k+2)}$

Nowyou can cancel (k+1). You should add to the final result that $k \ne -1$

5. I got up until the final step you have. After I find the common denominator of (k+1)(k+2), I add up the number and get (k+2) + 1. How did you get (k+2) - 1?

6. Originally Posted by jzellt
I got up until the final step you have. After I find the common denominator of (k+1)(k+2), I add up the number and get (k+2) + 1. How did you get (k+2) - 1?
the minus sign in front of the $\frac {k + 2}{(k + 1)(k + 2)}$ is what yielded the -1

note: $- \frac {k + 2}{(k + 1)(k + 2)} + \frac 1{(k + 1)(k + 2)} = \frac {1 - (k + 2)}{(k + 1)(k + 2)} = - \frac {(k + 2) - 1}{(k + 1)(k + 2)}$

got it?

can you continue?