The coefficient of the term independent of y in the expansion of y+1y23−y13+1−y−1y−y1210 is

y+1y23−y13+1−y−1y−y1210=y13+1y23−y13+1y23−y13+1−y12−1y12+1y12(y12−1)10=y13+1−y12+1y1210=y13+1−1−y−1210=y13−y−1210General term 10Cr(y13)10−r−y−12r= 10Cry10−r3(−1)ry−r2=(−1)r 10Cry20−2r−3r6=(−1)r 10Cry20−5r6For independent of y term power of y is equal to 0⇒20−5r6=0⇒5r=20⇒r=4Independent term =(−1)4 10C4=10×9×8×71×2×3×4=210

The coefficient of the term independent of y in the expansion of y+1y23−y13+1−y−1y−y1210 is

y+1y23−y13+1−y−1y−y1210=y13+1y23−y13+1y23−y13+1−y12−1y12+1y12(y12−1)10=y13+1−y12+1y1210=y13+1−1−y−1210=y13−y−1210General term 10Cr(y13)10−r−y−12r= 10Cry10−r3(−1)ry−r2=(−1)r 10Cry20−2r−3r6=(−1)r 10Cry20−5r6For independent of y term power of y is equal to 0⇒20−5r6=0⇒5r=20⇒r=4Independent term =(−1)4 10C4=10×9×8×71×2×3×4=210

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The coefficient of the term independent of y in the expansion of ${(\frac{y + 1}{y^{\frac{2}{3}} - y^{\frac{1}{3}} + 1} - \frac{y - 1}{y - y^{\frac{1}{2}}})}^{10}$ is

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112

105

210

answer is C.

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Detailed Solution

${(\frac{y + 1}{y^{\frac{2}{3}} - y^{\frac{1}{3}} + 1} - \frac{y - 1}{y - y^{\frac{1}{2}}})}^{10}$

$= {(\frac{(y^{\frac{1}{3}} + 1) (y^{\frac{2}{3}} - y^{\frac{1}{3}} + 1)}{y^{\frac{2}{3}} - y^{\frac{1}{3}} + 1} - \frac{(y^{\frac{1}{2}} - 1) (y^{\frac{1}{2}} + 1)}{y^{\frac{1}{2}} (y^{\frac{1}{2}} - 1)})}^{10}$

$= {(y^{\frac{1}{3}} + 1 - \frac{y^{\frac{1}{2}} + 1}{y^{\frac{1}{2}}})}^{10}$

$= {(y^{\frac{1}{3}} + 1 - 1 - y^{\frac{- 1}{2}})}^{10}$

$= {(y^{\frac{1}{3}} - y^{\frac{- 1}{2}})}^{10}$

General term $^{10} C_{r} {(y^{\frac{1}{3}})}^{10 - r} {(- y^{- \frac{1}{2}})}^{r}$

$\begin{array}{l} =^{10} C_{r} y^{\frac{10 - r}{3}} {(- 1)}^{r} y^{\frac{- r}{2}} \\ = {(- 1)}^{r}^{10} C_{r} y^{\frac{20 - 2 r - 3 r}{6}} \\ = {(- 1)}^{r}^{10} C_{r} y^{\frac{20 - 5 r}{6}} \end{array}$