Very Tricly Induction Question :D (1 Viewer)

[Smiley :D CvH]

hey guiis
got a tricky induction question ere which i cant figure out how to prove this ,.... can any one please help..... itd be greatly appreciated.... i put the picture up now.... fanks@!!!!

smiley

 

[Riviet]

Basically to do part a), substitute n=1,2,3 to find u₂,u₃,u₄ respectively, and remember to sub. in the previous result for each so for n=1, sub in u₁=1/2 into it. You should get 1/2, 2/3, 3/4, and by inspection the nth term would be n/(n+1). Now since you've already shown it's true for n=1, you can dive stright into the proof for n=k+1. I have to go watch a movie very soon, that's all I have time for. Hope that helped and good luck with the proof!
Riv.

 

[icycloud]

For n = 1:
LHS = u_1 = 1/2 (given)
RHS = k/(k+1) = 1/2 = LHS

Assume for n = k that:
u_k = k/(k+1)

Prove for n = k+1:

We must prove that u_(k+1) = (k+1)/(k+2)

Now, LHS = u_(k+1) = 1/(2-u_k) {given}
= 1/(2-(k/(k+1))) {using n = k assumption}
= 1/((2k+2-k)/(k+1))
= (k+1)/(k+2)
= RHS
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If it is true for n = k, then it is true for n = k+1. BUT it is true for n = 1, thus it is true for n = 2,3,4,... by the principles of mathematical induction.