IIT JEE , AIEEE Forum - Mathematics , Algebra

aggmegha90

The coeffient of the term independent of x in the expansion of

${\frac{x+1}{{x}^{\frac{2}{3}}-{x}^{\frac{1}{3}}+1}-\frac{x-1}{x-{x}^\frac{1}{2}}}^{10}$

aggmegha90

this whole expresion is raised to power 10

please reply!!
tell me some short method for it!

aggmegha90

please reply!

budokai_tenkaichi_returns

wat can i reply??

i just dont know how to do it

like u ..i m also waiting for d answer.

sandeepramesh

just a start (too lazy to post full soln :D)

The given expression is equivalent to ${x}^{\frac{1}{3}}+1 - (1-\frac{1}{{x}^{\frac{1}{2}}})$

whole power 10 ofc :)

aggmegha90

how? please explain!

asish

x+1=(x*1/3)*3+(1*1/3)*3

=(x*1/3+1)(x*2/3-x*1/3+1)

pantpranav

Nice question

asish

on solving we get (x*1/3+x*-1/2)*10

on solving we get term independent of x is 10c6=210

sandeepramesh

excuse me, i have already provided three fourths of the soln with the main idea :huh :huh And what do you mean by saying that u are waiting for the soln? :huh :huh

continuing from my soln (i hate to do this but still!)

So basically we need to find coeff of

${x}^{0} in ( x^{\frac{1}{3}} + x^{-\frac{1}{2}})^{10} = ?$

On computing we get :

coeff = 10C4 = 210

sriram.a

$[\frac{(x+1)(x^{1/3}+1)}{x+1}-\frac{x-1}{x^{1/2}(x^{1/2}-1)}]^{10}$

$[x^{1/3}+1-\frac{x^{1/2}+1}{x^{1/2}}]^{10}$

$[x^{1/3}+\frac{1}{x^{1/2}}]^{10}$

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now use binomial theorem and expand the equation

you will get answer as 210.

If u find my answer/satisfactory plzzzzzz rate meeeeeee.