Search any question & find its solution
Question:
Answered & Verified by Expert
The number of terms in the expansion of $\left(y^{1 / 5}+x^{1 / 10}\right)^{55}$, in which powers of $x$ and $y$ are free from radical signs are
Options:
Solution:
2672 Upvotes
Verified Answer
The correct answer is:
$\operatorname{six}$
$\operatorname{six}$
Given expansion is $\left(y^{1 / 5}+x^{1 / 10}\right)^{55}$ The general term is
$$
T_{r+1}={ }^{55} \mathrm{C}_r\left(y^{1 / 5}\right)^{55-r} \cdot\left(x^{\frac{1}{10}}\right)^r
$$
$T_{r+1}$ would free from radical sign if powers of $y$ and $x$ are integers.
i.e. $\frac{55-r}{5}$ and $\frac{r}{10}$ are integer.
$\Rightarrow r$ is multiple of 10 .
Hence, $r=0,10,20,30,40,50$
It is an A.P.
Thus, $50=0+(k-1) 10$ $50=10 k-10 \Rightarrow k=6$
Thus, the six terms of the given expansion in which $x$ and $y$ are free from radical signs.
$$
T_{r+1}={ }^{55} \mathrm{C}_r\left(y^{1 / 5}\right)^{55-r} \cdot\left(x^{\frac{1}{10}}\right)^r
$$
$T_{r+1}$ would free from radical sign if powers of $y$ and $x$ are integers.
i.e. $\frac{55-r}{5}$ and $\frac{r}{10}$ are integer.
$\Rightarrow r$ is multiple of 10 .
Hence, $r=0,10,20,30,40,50$
It is an A.P.
Thus, $50=0+(k-1) 10$ $50=10 k-10 \Rightarrow k=6$
Thus, the six terms of the given expansion in which $x$ and $y$ are free from radical signs.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.