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). ` `

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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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