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VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: the.watchman on February 05, 2010, 06:47:53 am

Title: the.watchman's weird q's thread
Post by: the.watchman on February 05, 2010, 06:47:53 am: I've got this question where we need to find the answer but not give working, so logically, it is quite possible to find.
However, I felt like proving it, but I got a little stuck.
(I'm using an induction method, if there's something better out there, then please tell me!)

Prove that $999...9^2 + 1999...9 = 10^{2n}$ , where 999...9 denotes n digits of 9 (so 1999...9 has n+1 digits)

Evidently $n\geq1$

When $n=1$ , $LHS = 9^2+19 = 100$ , $RHS = 10^{2\times1} = 100$

So $LHS = RHS$ , the equation holds for $n=1$

Assume that the equation is true for $n=k$ ,

$81(10^k+10^{k-1}+...+10+1)^2 + 10^k+1 + 9(10^k+10^{k-1}+...+10+1) = 10^{2k}$

I'm having trouble then with finding a way to sub the n=k equation into the n=k+1 equation
Can someone help me with that, or is there a better way of doing this???
Thanks!
Title: Re: the.watchman's weird q's thread
Post by: Neobeo on February 05, 2010, 06:53:46 am: Let $999...9 = 10^n-1$
Title: Re: the.watchman's weird q's thread
Post by: the.watchman on February 05, 2010, 07:03:48 am: Quote from: Neobeo on February 05, 2010, 06:53:46 am
Let $999...9 = 10^n-1$

Grrr ... why didn't I think of that!
Thanks! Now to prove...
Title: Re: the.watchman's weird q's thread
Post by: the.watchman on February 05, 2010, 07:11:05 am: Gee... that turns the proof into three steps...
Thanks heaps Neobeo!
I'm such an idiot...