Re: HSC 2013 4U Marathon
Here's my go at the question. I'll just do the left inequality and presume the right inequality is similar.
\frac{x^{n+1}-y^{n+1}}{x-y}=x^n+x^{n+1}y+...+xy^{n-1}+y^n
=y^n(1+\frac{x}{y}+\frac{x^2}{y^2}+...+\frac{x^n}{y^n})>y^n(n+1)
As \left ( \frac{x}{y} \right )^k>1
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You don't have to show that because if I was less than 2, the I(n-2) would be negative and therefore the equation won't work will just keep on going. I0 is the last value for reduction formulaes
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ax^3+bx^2+cx+d=0
Let roots be \frac{a}{c},a,ac
Using product of roots...
a^3=\frac{-d}{a}
a=\sqrt[3]{\frac{-d}{a}}
Subbing it back into the equation, as a is a root
a(\sqrt[3]{\frac{-d}{a}})^3+b(\sqrt[3]{\frac{-d}{a}})^2+c(\sqrt[3]{\frac{-d}{a}})+d=0
-d+b\left (...