http://img197.imageshack.us/img197/1866/image18tv.jpg

I would be very grateful if anyone can help :)

I think I have a way, but i'm a little dubious about it. It involves subtracting f(n) from f(n+1), and factorising out a 4

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- Sep 6th 2009, 03:25 AMLHSProof by Induction (2n+1)(7^n)-1=4x
http://img197.imageshack.us/img197/1866/image18tv.jpg

I would be very grateful if anyone can help :)

I think I have a way, but i'm a little dubious about it. It involves subtracting f(n) from f(n+1), and factorising out a 4 - Sep 6th 2009, 03:50 AMTaluivren
hint on the induction step:

$\displaystyle (2(n+1)+1)7^{n+1}-1 = (2n+1)7^{n+1} +2\cdot 7^{n+1} -1 = (2n+1)7^n + (2n+1)7^n\cdot 6 +2\cdot 7^{n+1} -1 =$

$\displaystyle = (2n+1)7^n -1 +(2n+1)7^n\cdot 6+ 2\cdot 7^n\cdot(3+4) = (2n+1)7^n -1 +(2n+1)7^n\cdot 6+ 7^n\cdot 6 +8\cdot 7^n = $

$\displaystyle = (2n+1)7^n -1 +(2n+2)7^n\cdot 6 +8\cdot 7^n $

can you follow from here? - Sep 6th 2009, 04:16 AMLHS
Ok, so what you have shown there is that:

$\displaystyle f(n+1) = f(n) + 4[7^n(3n+5)]$

So you are saying that if f(n+1) equals the original function, add a multiple of 4?

How does that prove that f(n) is a multiple of 4? - Sep 6th 2009, 04:21 AMTaluivren
do you know how proofs by induction work? Mathematical induction - Wikipedia, the free encyclopedia

1. base case: n=1 ... f(1) = 20 is divisible by 4, base case verified.

2. inductive step: suppose the statement holds for some n. show that the statement holds for n+1. And that is what i've done in the previous post. - Sep 6th 2009, 04:30 AMLHS
Sorry I'm being retarded. Yes I am meant to know how induction works :P I can do the simpler questions, this through me off as it had a n within the bracket as well as the power.

Yes, that makes perfect sense now! :D Thanks for your help, this has been bugging me!