induction help (1 Viewer)

[kr73114]

1+ 1/rt2 + 1/rt3 + ...+ 1/rt(n) > 2(rt(n+1) -1)

 

[iampeterr]

im assuming that rt stands for root? and i tried n values of 1,2 .. and they don't work ?

 

[binhvu24]

do it your seffff *untttttt

 

[Kingportable]

Um, dude don't be lazy and go re-wrtie this question properly if you want help. square root = sqrt(x)

 

[deswa1]

I'll give you a hint. If you still can't do it, I can post a full solution:
1. Assume it true for n=k
2. Test n=k+1
3. Sub in your assumption to get:
<a href="http://www.codecogs.com/eqnedit.php?latex=1@plus;\frac{1}{\sqrt2}@plus;...@plus;\frac{1}{\sqrt{k}}@plus;\frac{1}{\sqrt{k@plus;1}}>\frac{1}{\sqrt{k@plus;1}}@plus;2(\sqrt{k@plus;1}-1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?1+\frac{1}{\sqrt2}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\frac{1}{\sqrt{k+1}}+2(\sqrt{k+1}-1)" title="1+\frac{1}{\sqrt2}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\frac{1}{\sqrt{k+1}}+2(\sqrt{k+1}-1)" /></a>
4. Now put the RHS on the same denominator and try and go from there...