Doubt from binomial theorem

[Chirag_Hegde]

How to prove?
@Anindya_Sikdar @Viraam_Rao @Mayank_Chowdhary

[Mayank_Chowdhary]

It is the expansion of (1+x)^2n
Taking x= sixth root if unity I am getting something different answer
Also for the second and third proof simply substituting x=1 we get different answers I just don't know from where are they getting 3^n

[Chirag_Hegde]

Expansion is (1+x+x²)^n
Cropped the wrong part my bad.
Doubt is first one only btw
@Mayank_Chowdhary

[Mayank_Chowdhary]

Can you please send the whole question
I thought only a0 + a1x + a2x^2 ... Was given with the condition a0 = a2n ,. ....
So I assumed that it can be the expansion of (1+x)^2n

[Mayank_Chowdhary]

I think you can easily do it now !

[Azimuddin_Sheikh]

How to check whether to use (1+x)^n or (1+x^2+x)^n @Mayank_Chowdhary

[Mayank_Chowdhary]

That was given in the question

[Azimuddin_Sheikh]

Ohk

[Azimuddin_Sheikh]

@Mayank_Chowdhary bro I am getting this :

after doing something to this I am getting final answer as 1/6 (3^n +1+2 ) which is not the above pls check

[Mayank_Chowdhary]

@Azimuddin_Sheikh
Whatever you got in the pic is correct
Now we get

[Chirag_Hegde]

@Mayank_Chowdhary @Azimuddin_Sheikh

[Azimuddin_Sheikh]

@Chirag_Hegde bro

here mod means ,like take example 3 is 1 mod 2 , (I.e. 3 leaves remainder 1 when divided by 2 )

[Chirag_Hegde]

Thanks bro

[Mayank_Chowdhary]

[Azimuddin_Sheikh]

A slightly faster solution : didn't thought of -1 mod 7 thing .
We find that
2222^3\equiv3^3\equiv27\equiv-1\pmod7 Therefore,
2222^{5555}+5555^{2222}\equiv2222^{2222+3333}+(-2222)^{2222}\equiv2222^{2222}(-1)+2222^{2222}\equiv\boxed0\pmod7

[Chirag_Hegde]

What's the proof?

@Hari_Shankar sir
@Azimuddin_Sheikh

[Hari_Shankar]

Use the well known theorem that 2< \left(1+\dfrac{1}{n} \right)^n<3

Let a_n = \left( \dfrac{n}{3}\right)^n; b_n=n!, c_n = \left( \dfrac{n}{2}\right)^n

\dfrac{a_{n+1}}{a_n} = \left(\dfrac{n+1}{n}\right)^n \times \dfrac{n+1}{3}<n+1 = \dfrac{b_{n+1}}{b_n}

i.e. the sequence a_n grows slower than b_n and clearly a_1<b_1. Hence a_n<b_n \ \forall \ n \in \mathbb{N}

[Chirag_Hegde]

Did this easily by expanding it .
But what if numbers got bigger? Is there a more general method?

[Yash_Srivastava2]

take (√2 -1)^6 add let it be x and make some inequality you will get it and remember taking (√2+1)^6 as [y] +{y} then you will only get beneficial inequality

[Azimuddin_Sheikh]

@Yash_Srivastava2 bro by this method we will get x+{y} =1 , and (1-{y} )( [y] + {y}) = 1, how will this be beneficial ??