Help needed..

If the rth{ r }^{ th } term in the expansion of (2x+1xm)n,(nϵN){ \left( 2x+\frac { 1 }{ { x }^{ m } } \right) }^{ n },(n\epsilon N) is independent term of xx and rth{ r }^{ th } term in expansion of (1x42xn6)m,(mϵN){ \left( \frac { 1 }{ { x }^{ 4 } } -2{ x }^{ n-6 } \right) }^{ m },(m\epsilon N) is independent term of xx, then minimum possible value of [m2+n29]\left[ \frac { { m }^{ 2 }+{ n }^{ 2 } }{ 9 } \right] is? (where [.]\left[ . \right] represents greatest integer function)

What I have done is write down the general term in both expansions and equated the powers of xx to zero in both cases and tried to connect them...But it goes nowhere..

Note by Anandhu Raj
4 years, 5 months ago

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@Sandeep Bhardwaj

Anandhu Raj - 4 years, 5 months ago

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When you try to connect, you got two equations in m,n and rm,n \ and \ r then substitute the value of rr in terms of m,nm,n from one equation in the second equation and simplify.

On simplifying you will get :

4m2+4m=n22n4m^2+4m=n^2-2n

4m2n2=2(2m+n)4m^2-n^2=-2(2m+n)

(2m+n)(2mn)=2(2m+n)(2m+n)(2m-n)=-2(2m+n)

So, either 2m+n=02m+n=0 or 2mn=22m-n=-2.

The first one is not possible because m,nNm,n \in \mathbb{N} .

So 2mn=22m-n=-2

2m=n22m=n-2.

Now, we'll focus on the value of nn :

From the second equation, it is sure that n6n \geq 6.

If n=6,    m=2n=6 , \implies m=2 (Min. Permissible values)

Hence the answer comes to be 44 .

edited !

Hope you'll enjoy it !

Sandeep Bhardwaj - 4 years, 5 months ago

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Why n>6?n > 6?

n=6n = 6 , m=2m = 2 also satisfies the condition And also gives the minimum value of 4.

Clarification:- check the last term of second equation with n= 6( power of 1/x41/x^4 equals 0).

Krishna Sharma - 4 years, 5 months ago

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You're right. I made a mistake by neglecting n=6. Thanks !

Sandeep Bhardwaj - 4 years, 5 months ago

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Thanks sir :)

Anandhu Raj - 4 years, 5 months ago

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