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Sum

If the term free from x in the expansion of `(sqrt(x) - k/x^2)^10` is 405, find the value of k.

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#### Solution

The given expression is `(sqrt(x) - k/x^2)^10`

General term `"T"_(r + 1) = ""^n"C"_r x^(n - r) y^r`

= `""^10"C"_r (sqrt(x))^(10 - r) ((-k)/x^2)^r`

= `""^10"C"_r (x)^((10 - r)/2) (-k)^r (1/x^(2r))`

= `""^10"C"_r (x)^((10 - r)/2 - 2r) (-k)^r`

= `""^10"C"_r (x)^((10 - r - 4r)/2) (- k)^r`

= `""^10"C"_r (x)^((10 - 5r)/2) (- k)^r`

For getting term free from x

`(10 - 5r)/2` = 0

⇒ r = 2

On putting the value of r in the above expression

We get `""^10"C"_2 (-k)^2`

According to the condition of the question, we have

`""^10"C"_2 k^2` = 405

⇒ `(10*9)/(2*1) k^2` = 405

⇒ 45k^{2} = 405

⇒ k^{2} = `405/45` = 9

∴ k = `+- 3`

Hence, the value of k = ±3

Concept: General and Middle Terms

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