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Binomial Proof (1 Viewer)
- Thread starter sizzly
- Start date
RealiseNothing
what is that?It is Cowpea
I'm confused as to how this is possibly a binomial proof?
Squar3root
realest nigga
how dafaq did you see that? lel and (iii) should come immediately from the recursion relation in (i)$ \\ \\ $Proceeding$ \\ \\ $(i)$ \ I_{n+1} = \int_1^2 \left(1 - \frac{1}{x} \right)^{n+1} \ dx \\ \\ u = \left(1 - \frac{1}{x} \right)^{n+1} \ , \ du = \frac{(n+1)}{x^2} \left(1 - \frac{1}{x} \right)^{n} \\ dv = 1 \ , \ v = x \\ \\ \therefore \ I_{n+1} = \frac{1}{2^n} - (n+1)\int_1^2 \frac{1}{x} \left(1 - \frac{1}{x} \right)^{n} \\ \\ $Notice however$ \ \frac{1}{x} \left(1 - \frac{1}{x} \right)^n = \left( 1 - \frac{1}{x} \right)^n - \left( 1 - \frac{1}{x} \right)^{n+1} \\ \\ \therefore \ I_{n+1} = \frac{1}{2^n} - (n+1) (I_{n} - I_{n+1}) \\ \\ \Rightarrow \ \frac{I_{n+1}}{n+1} = \frac{I_n}{n} - \frac{1}{n(n+1)2^n} )
RealiseNothing
what is that?It is Cowpea
It's a common integration trick, it was used in our 2013 HSC exam too for question 13)a).
If I'm not mistaken, OP's question is the last question of the 2013 independent paper?