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  • If the term free from x is the expansion of square root of x minus k divided by x square Exponents of 10 is 405, then find the value of k.

if the term free from x is the expansion of  \left ( \sqrt{x}-\frac{k}{x^{2}} \right )^{10} is 405, then find the value of k.

Answers (1)

\left ( \sqrt{x}-\frac{k}{x^{2}} \right )^{10}................... (Given)

T_{r+1}=^{10}\textrm{C}_{r}\left ( \sqrt{x} \right )^{10-r}\left ( -\frac{k}{x^{2}} \right )^{r}  ……. (from standard formula of Tr+1)

          

T_{r+1}=^{10}\textrm{C}_{r} x ^{\frac{10-5r}{2}}\left ( -k \right )^{r} …….. (i)

Now, for x,

(10 – 5r)/2 = 0

  → r = 2

Substituting the value of r in eq (i),

T_{2+1}=^{10}\textrm{C}_{2} \left (-k \right )^{2}

            = 405

\frac{10\times 9\times 8!}{2!\times 8!}\left ( k \right )^{2}=405

45k^{2}=405

Thus k2 = 9

Thus, k = ±3

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