# 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 $x$ to zero in both cases and tried to connect them...But it goes nowhere..

Note by Anandhu Raj
6 years, 1 month ago

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- 6 years, 1 month ago

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

On simplifying you will get :

$4m^2+4m=n^2-2n$

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

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

So, either $2m+n=0$ or $2m-n=-2$.

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

So $2m-n=-2$

$2m=n-2$.

Now, we'll focus on the value of $n$ :

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

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

Hence the answer comes to be $4$.

edited !

Hope you'll enjoy it !

- 6 years, 1 month ago

Why $n > 6?$

$n = 6$, $m = 2$ 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 $1/x^4$ equals 0).

- 6 years, 1 month ago

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

- 6 years, 1 month ago

Thanks sir :)

- 6 years, 1 month ago