# 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
3 years, 1 month ago

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- 3 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 !

- 3 years, 1 month ago

Thanks sir :)

- 3 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).

- 3 years, 1 month ago

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

- 3 years, 1 month ago

Comment deleted May 31, 2015

I got this question from an old question paper. Going by the key code the answer is 4(but no steps given)

- 3 years, 1 month ago

Check I've posted a brief solution too.

- 3 years, 1 month ago

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