Doubt from binomial theorem

[Yash_Srivastava2]

@Azimuddin_Sheikh

[Azimuddin_Sheikh]

Awesome @Yash_Srivastava2 bro !!! Didn't thought of it

[Azimuddin_Sheikh]

Got inspiration from u @Yash_Srivastava2 bro : another way : Quantity is (3+2\sqrt{2})^3

Let a_n=(3+2\sqrt{2})^n+(3-2\sqrt{2})^n so that a_0=2, a_1=6 and a_{n+2}=6a_{n+1}-a_n

We get a_2=6^2-2=34 and a_3=6\times 34-6=198

And since (3-2\sqrt{2})^3\in(0,1), we get \left\lfloor(3+2\sqrt{2})^3\right\rfloor=a_3-1

\boxed{197}

[Yash_Srivastava2]

an+2=6an+1−an

Bro can you tell me how did you arrive at this result by intuition?........now i know i can prove it....but how you thought that such pattern will follow @Azimuddin_Sheikh

[Azimuddin_Sheikh]

@Yash_Srivastava2 bro , If a_n=\alpha u^n+\beta v^n, it is classical (and easy to check) that
a_{n+2}=(u+v)a_{n+1}-uv a_n I remember it because there was once a question of this type in previous year kvpy

[Yash_Srivastava2]

do you know where i can find such things like this one?

[Azimuddin_Sheikh]

This thing is in recurrence relation chapter , I studied when I was very interested in functional equation and number theory stuff. U can find this in net also @Yash_Srivastava2 bro

[Chirag_Hegde]

Doubt in both
@Azimuddin_Sheikh @Mayank_Chowdhary

[Azimuddin_Sheikh]

First one : Let S be the sum.Then S = (n-1) \cos \left( \dfrac{2\pi}{n} \right) + (n-2) \cos \left( \dfrac{4\pi}{n} \right) + \cdots + \cos \left( \dfrac{2(n-1)\pi}{n} \right) .Writing S in reverse order we get S = \cos \left( \dfrac{2(n-1)\pi}{n} \right) + 2 \cos \left( \dfrac{2(n-2)\pi}{n} \right) + \cdots + (n-1) \cos \left( \dfrac{2\pi}{n} \right).Adding both and also using \cos ( 2\pi - x) = \cos (x) to cancel out the terms we get 2S = n \left( \cos \left( \dfrac{2\pi}{n} \right) + \cos \left( \dfrac{4\pi}{n} \right) + \cdots + \cos \left( \dfrac{2(n-1)\pi}{n} \right) \right).This is just an AP of the involved angles and using formula for AP we get S = \dfrac{ n \sin \left( \dfrac{(n-1)\pi}{n} \right) \cos \left( \dfrac{2 \pi}{n} + \dfrac{(n-2) \dfrac{2 \pi}{n}}{2} \right) }{ 2 \sin \left( \dfrac{\pi}{n} \right)} = \dfrac { n \cos \pi }{2} = \boxed{ \dfrac{-n}{2} }

[Tushar_Rathore]

2nd one