If the term free from x in the expansion of -x-k-x210 is 405 - find the value of k-

Let (r+1)^th term, in the expansion of ( √(x) k/x^2 )^10, be equal to T_r+1=^10C_r (√(r))^10 r ( k/x^2 )^r =^10C_r x^5 5r/2( l)^r....(i) If T_r+1 is independen ...

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Standard XII

Mathematics

Term Independent of x

If the term f...

Question

If the term free from x in the expansion of ${(\sqrt{x} - \frac{k}{x^{2}})}^{10}$ is 405, find the value of k.

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Solution

Let $(r + 1)^{t h}$ term, in the expansion of ${(\sqrt{x} - \frac{k}{x^{2}})}^{10}$ , be equal to $T_{r + 1} =^{10} C_{r} (\sqrt{r})^{10 - r} {(\frac{- k}{x^{2}})}^{r}$

$=^{10} C_{r} x^{5 - \frac{5 r}{2}} (- l)^{r} . . . . (i)$

If $T_{r + 1}$ is independent of x, then $5 - \frac{5 r}{2} = 0$

$\Rightarrow r = 2$

Putting r =2 in(i), we obtain

$T_{3} =^{10} C_{2} (- k)^{2} = 45 k^{2}$

But it is given that the value of the term free from x is 405

$∴ 45 k^{2} = 405 \Rightarrow k^{2} = 9 \Rightarrow k = \pm 3$

Hence, the value of k is $\pm 3.$

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If the term free from x in the expansion of (√x−kx2)10 is 405, find the value of k.

If the term free from x in the expansion of ${(\sqrt{x} - \frac{k}{x^{2}})}^{10}$ is 405, find the value of k.