If the coefficient of x in the expansion of x2-k-x5 is 270 - then k-A- 2B- 1C- 3D- 4

The correct option is C 3T_r+1=^5C_r(x^2)^5 r ( k/x )^r For coefficient of x, 10 – 2r – r = 1⇒ r=3 Hence, T_3+1=^5C_r(x^2)^5 3 ( k/x )^3 According to question, ...

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Standard XII

Mathematics

Expansions of the Form (1+x)^(-n) and (1-x)^(-n)

If the coeffi...

Question

If the coefficient of x in the expansion of ${(x^{2} + \frac{k}{x})}^{5}$ is 270, then k=

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Solution

The correct option is C 3
$T_{r + 1} =^{5} C_{r} (x^{2})^{5 - r} {(\frac{k}{x})}^{r}$
For coefficient of x, 10 – 2r – r = 1 $\Rightarrow r = 3$
Hence, $T_{3 + 1} =^{5} C_{r} (x^{2})^{5 - 3} {(\frac{k}{x})}^{3}$
According to question, $10 k^{3} = 270 \Rightarrow k = 3$

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The correct option is C 3Tr+1=5Cr(x2)5−r(kx)r For coefficient of x, 10 – 2r – r = 1⇒r=3 Hence, T3+1=5Cr(x2)5−3(kx)3 According to question, 10k3=270⇒k=3

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The correct option is C 3
$T_{r + 1} =^{5} C_{r} (x^{2})^{5 - r} {(\frac{k}{x})}^{r}$
For coefficient of x, 10 – 2r – r = 1 $\Rightarrow r = 3$
Hence, $T_{3 + 1} =^{5} C_{r} (x^{2})^{5 - 3} {(\frac{k}{x})}^{3}$
According to question, $10 k^{3} = 270 \Rightarrow k = 3$