Question

Write down and simplify:
The $5^{th}$ term of ${(\frac{x^{\frac{3}{2}}}{a^{\frac{1}{2}}}-\frac{y^{\frac{5}{2}}}{b^{\frac{3}{2}}})}^8$.

• A. $\frac{60x^6{y}^5}{a^2{b}^6}$
• B. $\frac{70x^6{y}^{10}}{a^3{b}^5}$
  
• C. $\frac{70x^5{y}^5}{a^2{b}^6}$
  
• D. $\frac{70x^6{y}^{10}}{a^2{b}^6}$
  

Answer 1

Answer: (D)

$\frac{70x^6{y}^{10}}{a^2{b}^6}$

The $r^{th}$ term in expansion of $(p+q{)}^n$ is given by ${}^n{C}_{r-1}{p}^{n+1-r}{q}^{r-1}$

$\therefore$ in the given expression

$p=\frac{x^{\frac{3}{2}}}{a^{\frac{1}{2}}},q=-\frac{y^{\frac{5}{2}}}{b^{\frac{3}{2}}},r=5\&n=8$

$\therefore$ the $5^{th}$ term will be ${}^8{C}_4{\left(\frac{x^{\frac{3}{2}}}{a^{\frac{1}{2}}}\right)}^4{(-\frac{y^{\frac{5}{2}}}{b^{\frac{3}{2}}})}^4=\frac{70x^6{y}^{10}}{a^2{b}^6}$