find-the-fourth-term-in-the-expansion-of-x-y-2-y-x-6-

Question Number 55301 by Otchere Abdullai last updated on 20/Feb/19

f i n d t h e f o u r t h t e r m i n t h e e x p a n s i o n

o f {(\frac{\sqrt{x}}{y^{2}} - \frac{y}{\sqrt{x}})}^{6}

Commented by mr W last updated on 21/Feb/19

{(\frac{\sqrt{x}}{y^{2}} - \frac{y}{\sqrt{x}})}^{6}

= {(\frac{\sqrt{x}}{y^{2}})}^{6} {(1 - \frac{y^{3}}{x})}^{6}

= \frac{x^{3}}{y^{12}} {(1 - \frac{y^{3}}{x})}^{6}

4 t h t e r m = \frac{x^{3}}{y^{12}} \times C_{3}^{6} \times {(- \frac{y^{3}}{x})}^{3}

= \frac{x^{3}}{y^{12}} \times 20 \times (- \frac{y^{9}}{x^{3}})

= - \frac{20}{y^{3}}

Commented by Otchere Abdullai last updated on 21/Feb/19

G o d b l e s s y o u P r o f W

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Feb/19

r + 1 t h t e r m \to {n c}_{r} {(a)}^{n - r} {(b)}^{r}

s o 4 t h t e r m \to 6 c_{3} {(\frac{\sqrt{x}}{y^{2}})}^{6 - 3} {(\frac{- y}{\sqrt{x}})}^{3}

= \frac{6!}{3! \times 3!} \frac{{(x)}^{\frac{3}{2}}}{{(y)}^{6}} \times (- 1) \times \frac{y^{3}}{{(x)}^{\frac{3}{2}}}

= \frac{- 20}{y^{3}}

Commented by Otchere Abdullai last updated on 21/Feb/19

t h a n k y o u p r o f T a n m a y