# The total no of term free from radical signs in the expansion of (y^{1/5}+x^{1/10})^55

Arun Kumar IIT Delhi
10 years ago
Hello Student,
The general term in the expansion of (y^{1/5}+x^{1/10})^55
is given by
$\\T_{r+1}=\binom{55}{r}(y^{1/5})^{55-r}(x^{1/10})^r \\=\binom{55}{r}y^{11-r/5}x^{r/10}$
=>r/5,r/10 should be integers
=>r=0,10,20,30,40,50
=>6 terms
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
19 Points
7 years ago
(y1/5 +x1/10)55
General term= 55Cr(y1/5)55-r(x1/10)r
=55Cr(y)11-r/5(x)r/10
r/5 and r/10 should be integers
so the first value for r = 0.
after this add the lcm of 5 and 10 i.e, 10 untill the value of r become equal to the value of “n” or smaller than it , i.e 0,10,20,30,40,50
so there are 6 terms.
Abhishek
14 Points
6 years ago
It`s a trick which can find in a easy way in a couple of seconds. (X^1/5 +y^1/10)^55 No of terms =[n/LCM of p and q] + 1 Where p=1/5 and q=1/10 and [...] Is greatest integer factor so now =[55/10] +1=[5.5]+1=5+1=6 solved gif is a mathematical term