Re: Induction problems

From: Steven (sgottlieb60_at_hotmail.com)
Date: 10/31/04 

Date: Sun, 31 Oct 2004 02:34:21 GMT

"Neil L." <imstr8trippin@yahoo.com> wrote in message
news:-YOdnWnM0afwQx7cRVn-pw@rogers.com...
> Please someone out there explain this to me,
>
> I just started working on Induction problems and can't seem to figure this
> out....
>
> Here's the whole problem. Prove that 3^n > 20n, for each integer n >= 4.
>
> so, n=4, 3^4 = 81 and 20n = 80, so 3^n > 20n
>
> Now, assume induction hypothesis 3^k > 20k, for each integer k >= 4. We
> must prove 3^(k+1) > 20(k+1) We have:
>
> 3^(k+1) = 3(3^k)
> 3^(k+1) > 3(20k) (by the induction hypothesis)
> 3^(k+1) > 20k + 40k
> 3^(k+1) > 20k + 20 (since k >= 4 > 1/2)

Since k>4 then multiplying both sides by 40 gives us 40k>160 > 20 (since 160
is greater than 20)
> 3^(k+1) > 20(k+1)
>
> The 2nd last step is where my issue comes in. I DONT GET IT!!!
>
> Can someone please explain? Thank you.
>
>