# The expansion of (2x^(2)-1/sqrt(x))^(n) where n is a positive integer,has a term that is independent of x.Find the smallest value of n.Thanks!   (Using the binomial theorem if possible). 

tiburtius | High School Teacher | (Level 2) Educator

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General term in expansion (using binomial theorem) would be

(-1)^k((n),(k))(2x^2)^(n-k)(1/(sqrt x))^k

However, we are interested only in term that is independent of x so we will disregard everything else. If the term is independent of x it means that both powers of x must be the same i.e.

x^(2(n-k))=(sqrt x )^k

x^(2(n-k))=x^(k/2)    because sqrt x=x^(1/2)

2(n-k)=k/2

4(n-k)=k

4n-4k=k

4n=5k

n=(5k)/4

In order to find the smallest n we need to determine the smallest k>0 for which n is a whole number. Obviously we get that for k=4 (for k=1,2,3  n is not a whole number).

n=(5\cdot4)/4

n=5

The smallest value of n is 5.

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