Question

The $7^{\text{th}}$ term in ${(\frac{1}{y}+y^2)}^{10},$ when expanded in descending power of $y,$ is

• A. $210y^8$
• B. $\frac{210}{y^8}$
• C. $\frac{210}{y^2}$
• D. $210y^2$

Answer 1

The correct option is D $210y^2$

When ${(\frac{1}{y}+y^2)}^{10}$ is expanded, the power of $y$ go on increasing as the terms proceed. Hence it is expanded in ascending powers of $y.$

But, we have to expand in descending powers of $y.$
So, ${(y^2+\frac{1}{y})}^{10}$ when expanded will be in descending power of $y.$
$\therefore T_7={}^{10}{C}_6(y^2{)}^4{\left(\frac{1}{y}\right)}^6=210y^2$

Alternate Solution:
Given : ${(\frac{1}{y}+y^2)}^{10}$
General term is
$T_{r+1}={}^{10}{C}_r{\left(\frac{1}{y}\right)}^{10-r}(y^2{)}^r\Rightarrow T_{r+1}={}^{10}{C}_r{y}^{3r-10}$
As $r$ increases, power of $y$ also increases.
$\therefore 7^{\text{th}}$ term when expanded in descending power of $y$ is equal to the $5^{\text{th}}$ term when expanded in ascending power of $y.$
So, $T_5={}^{10}{C}_4{y}^{3\times 4-10}=210y^2$