Grade 11Algebra

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

MYQUESTIONS

12 Years agoGrade 11

3 Answers

Arun Kumar

Approved Tutor Answer12 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
Askiitians Faculty

yadav shiykhar

9 Years ago

(y^1/5+x^1/10)⁵⁵

General term=⁵⁵C_r(y^1/5)^55-r(x^1/10)^r

=⁵⁵C_r(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

8 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