Doubt from binomial theorem

[Yash_2019]

@Azimuddin_Sheikh

[Azimuddin_2019]

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

[Azimuddin_2019]

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_2019]

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_2019]

@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_2019]

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

[Azimuddin_2019]

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_2019]

Doubt in both
@Azimuddin_Sheikh @Mayank_Chowdhary

[Azimuddin_2019]

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_2019]

2nd one

[Chirag_2019]

@Mayank_Chowdhary @Viraam_Rao @Tushar_Rathore
Both

[Mayank_2019_1]

For the first is the answer zero ?
I did by mod theory... Will have to think about binomial one

[Rafe_2019]

@Chirag_Hegde
Is the answer zero ?
Write (13)^99-(19)^93 as (12+1)^99-(18+1)^93 and then proceed.

[Chirag_2019]

Any method is appreciated bro

[Mayank_2019_1]

For the second one
I know it is a bit shabby but I did some too silly mistakes ...
If you want I will write it more clearly bro

[Rafe_2019]

@Chirag_Hegde
Did you get my method or should I send solution ?

[Chirag_2019]

I'm trying out your method I'll let you know

[Mayank_2019_1]

For the first one
See we know 162=2.3^4 Now if the given number is divisible by 81 and is even we can say that the given number is divisible by 162
So 13^99 is odd 19^93 is also odd So the given number is even (as odd-odd=even)
Now 13=1mod(3) and 19=1mod(3)

Also $13=(12+1)=1(mod3) so 13^{99} = (12+1)^{99} = 1 (mod3) = (12.99+1)(mod81)=55(moid81)
In the last step i took the expansion of 12+1 and considered only the last two terms as rest all terms are divisible by 81
Similarly for 19 we do
19^{93} = (18+1){^93}= (18.93 +1 ) = 55(mod 81)
So subtracting both we get zero
PS: I guess I have seen the same problem somewhere else too ...Unable to recall

[Chirag_2019]

I wrote the terms which aren't directly divisible but it is going long
Is it being simplified ?
@Rafe_Moin1

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PS I posted this a while back But just realised net was off

[Mayank_2019_1]

https://brainly.in/question/482232
Just remembered it was an RMO question !
https://www.quora.com/How-do-I-prove-that-13-99-19-93-is-divisible-by-162
See the first answer on Quora for solution using binomial