1. ## Induction proof :D

Can you help me with this one? Take it slow please....

Show by induction: 2+2^2+2^3+.....+2^n=2*(2^n-1)

2. Hello, Neffets!

LaTeX isn't back yet . . . Hope you can read my notation . . .

Show by induction: .2 + 2^2 + 2^3 + . . . + 2^n .= .2(2^1 - 1}

Verify S(1): .2 .= .2(2^1 - 1) . . . yes!

Assume S(k): .2 + 2^2 + 2^3 + . . . + 2^k .= .2(2^k - 1)

. . 2 + 2^2 + 2^3 + . . . + 2^k + 2^{k+1} .= .2(2^k - 1) + 2^{k+1}

The right side is: .2^{k+1} - 2 + 2^{k+1} .= .2·2^{k+1} - 2 .= .2(2^{k+1} - 1)

And we have: .2 + 2^2 + 2^3 + . . . + 2^{k+1} .= .2(2^{k+1} - 1)

. . There!

3. It's a bit tricky without the LaTex, but i did understand what you wrote.. Although where did you get this from:
The right side is: .2^{k+1} - 2 + 2^{k+1}
Thanks for all of your help...

4. Originally Posted by Neffets...:P
It's a bit tricky without the LaTex, but i did understand what you wrote.. Although where did you get this from:

Thanks for all of your help...
Look at the previous line, on the right hand side of the "=" sign you
have:

2(2^k - 1) + 2^{k+1},

Now multiply the first bracket out to get:

(2^{k+1} - 2) + 2^{k+1}=2^{k+1} - 2 + 2^{k+1}

which is what you asked about, which Soroban then simplifies a bit
and puts back equal to the left hand side back at the earlier equation
where he had picked up the right hand side.

RonL

5. Of course...

Thnx Soroban and Captain Black!

6. Originally Posted by CaptainBlack
Fast cars, fast women, fast algorithms... what more could a man want?
-- Joe Mattis
RonL
Rolex watch.