My Math Forum hello all :) I don't know where I can put algorithms and please don't block this ))

 Algebra Pre-Algebra and Basic Algebra Math Forum

 November 16th, 2014, 12:52 PM #1 Newbie   Joined: Nov 2014 From: georgia Posts: 1 Thanks: 0 hello all :) I don't know where I can put algorithms and please don't block this )) F(4) = 17 F(n) = 2F(n-1) + 2n there is a little algorithm and can you solve it .. I solved it and in my opinion it is F(n) = F(n) = 2^(n-4)*F4 + (5+n)(n-4) that one, but someone says that this is not true. Can you solve this and explain why?? Thanks in advance. Last edited by skipjack; November 16th, 2014 at 02:40 PM.
 November 16th, 2014, 02:50 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,868 Thanks: 1833 That's not quite right. The solution is F(n) = (F(4) + 12)2^(n-4) - 2n - 4.
November 22nd, 2014, 12:45 AM   #3
Newbie

Joined: Nov 2014
From: HK

Posts: 3
Thanks: 1

I don't get it

Quote:
 F(4) = 17 F(n) = 2F(n-1) + 2n
Shouldn't it be something like this?

17 = 2F(4-1) + 2(4)
17 = 6F + 8
9 = 6F
F = 3/2

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