Help needed..

If the \({ r }^{ th }\) term in the expansion of \({ \left( 2x+\frac { 1 }{ { x }^{ m } } \right) }^{ n },(n\epsilon N)\) is independent term of \(x\) and \({ r }^{ th }\) term in expansion of \({ \left( \frac { 1 }{ { x }^{ 4 } } -2{ x }^{ n-6 } \right) }^{ m },(m\epsilon N)\) is independent term of \(x\), then minimum possible value of \(\left[ \frac { { m }^{ 2 }+{ n }^{ 2 } }{ 9 } \right] \) 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..

#Combinatorics

Note by Anandhu Raj
6 years ago

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Comments

@Sandeep Bhardwaj

Anandhu Raj - 6 years ago

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=n2−2n4m^2+4m=n^2-2n

4m2−n2=−2(2m+n)4m^2-n^2=-2(2m+n)

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

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

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

So 2m−n=−22m-n=-2

2m=n−22m=n-2.

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

From the second equation, it is sure that n≥6n \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 - 6 years 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 - 6 years ago

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

Sandeep Bhardwaj - 6 years ago

Thanks sir :)

Anandhu Raj - 6 years ago
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