Question

In the expansion of ${(\sqrt[5]{53}+\sqrt[3]{37})}^{36}$, the integral terms are

• A. $T_6,T_{21},T_{36}$
• B. $T_7,T_{22},T_{37}$
• C. $T_7,T_9,T_{11}$
• D. $T_5,T_{20},T_{35}$

Answer 1

The correct option is B $T_7,T_{22},T_{37}$

For $(\sqrt[5]{53}+\sqrt[3]{37}){}^{36}$
Number of rational terms is
$\frac{36}{L.C.M(5,3)}+1$
$=\frac{36}{15}+1$
$=2.4+1$ but we consider the integral value of $\frac{36}{15}$
Hence, number of ration terms is $2+1=3$
Consider the identity
$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
where T is the term no.
Here, $n=36$
$T_{r+1}={}^{36}{C}_r{3}^{\frac{36-r}{5}}{7}^{\frac{r}{3}}$
For $r=6,21,36$ we get rational terms with term numbers $T_7,T_{22},T_{37}$ respectively
Hence, answer is $B.$