Understanding Induction: Solving a Challenging Step

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i can't understand this step in this induction

http://img383.imageshack.us/img383/5677/99884212ia5.gif
 
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Because if an + L - e < an+1,

then an + 2(L - e) < an+1 + L - e

but an+1 + L - e < an+2

so an + 2(L - e) < an+1 + L - e < an+2 :smile:

Now try it for the other half, and write it out properly as an induction proof. :wink:
 
in the total case of K
we are given some expression
and we presume that its true
so to prove the K+1 case

from where did they use the left side development
it seems that the are using the given expression itself to proove it
?
http://img149.imageshack.us/img149/6359/10189365rk3.gif
 
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this is the expanded proove
http://img220.imageshack.us/img220/1912/13164747xs4.gif
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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