Question

The coefficient of $\frac{1}{y^2}$ in ${(y+\frac{c^3}{y^2})}^{10}$ is $3360$. If the value of $y=2$, then

• A. $c=\sqrt[3]{32}$
• B. greatest term will be $2880$
• C. $5^{\text{th}}$ term will be coefficient of $y^{-2}$
• D. $3^{\text{rd}}$ term will have greatest numerical value

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $c=\sqrt[3]{32}$
B greatest term will be $2880$
C $5^{\text{th}}$ term will be coefficient of $y^{-2}$
D $3^{\text{rd}}$ term will have greatest numerical value

The general term of the given expansion is
$T_{r+1}={}^{10}{C}_r{y}^{10-r}{\left(\frac{c^3}{y^2}\right)}^r\Rightarrow T_{r+1}={}^{10}{C}_r{y}^{10-3r}{c}^{3r}$
Now for $y^{-2}$ term
$10-3r=-2\Rightarrow r=4$
Hence $5^{\text{th}}$ term will have $y^{-2}$
Now the coefficient will be
$3360={}^{10}{C}_4{c}^{12}\Rightarrow 210c^{12}=210\times 16\Rightarrow c^3=2$
Now for numerically greatest term
$\left|\frac{T_{r+1}}{T_r}\right|\ge 1\Rightarrow |\frac{11-r}{r}\cdot \frac{c^3}{y^3}|\ge 1\Rightarrow \left|\frac{11-r}{4r}\right|\ge 1\Rightarrow r\le \frac{11}{5}$
Hence for $r=2$ the term will be greatest
The greatest term will be
$T_{2+1}={}^{10}{C}_2{y}^{10-6}{c}^6=2880$