# Induction proof :D

• Sep 16th 2006, 03:29 AM
Neffets...:P
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)

• Sep 16th 2006, 04:58 AM
Soroban
Hello, Neffets!

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

Quote:

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!

• Sep 16th 2006, 05:14 AM
Neffets...:P
It's a bit tricky without the LaTex, but i did understand what you wrote.. Although where did you get this from:
Quote:

The right side is: .2^{k+1} - 2 + 2^{k+1}
Thanks for all of your help...
• Sep 16th 2006, 06:36 AM
CaptainBlack
Quote:

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
• Sep 16th 2006, 07:00 AM
Neffets...:P
Of course...

Thnx Soroban and Captain Black!
• Sep 16th 2006, 08:01 AM
OReilly
Quote:

Originally Posted by CaptainBlack
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RonL

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