Question

Find the term of the expansion of ${(\sqrt[3]{3x}-\frac{1}{\sqrt{x}})}^{15}$ which does not contain $x$.

• A. $5$th
• B. $6$th
• C. $7$th
• D. $8$th

Answer 1

The correct option is C $7$th

The greatest term of ${(r+1)}^{th}$ term in the expansion of ${(a+b)}^n$ is given by,

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

Here, $n=15,a=\sqrt[3]{3x},b=\frac{1}{\sqrt{x}}$

${}^{15}{C}_r{\sqrt[3]{3x}}^{15-r}(-\frac{1}{\sqrt{x}}{)}^r$=$T_{r+1}$

${}^{15}{C}_r{x}^{\frac{15-r}{3}}(-\frac{1}{\sqrt{x}}{)}^r$=$T_{r+1}$

${}^{15}{C}_r(-1{)}^r\frac{x^{\frac{15-r}{3}}}{x^{\frac{r}{2}}}$=$T_{r+1}$

${}^{15}{C}_r(-1{)}^r{x}^{\frac{15-r}{3}-\frac{r}{2}}$=$T_{r+1}$.

So we need to find the value for $x$ such that,

$\frac{15-r}{3}-\frac{r}{2}=0$ ,only then that term will not contain $x$

$30-2r-3r=0$

$30-5r$

$r=6$

$T_{6+1}=T_7={}^{15}{C}_6(-1{)}^6{x}^0={}^{15}{C}_6$

Therefore $7^{th}$ term will not contain $x$.