Question

the number of terms in the expansion of (y^(1/5)+x^(1/10))^55,in which the powers of x and y are free from radical signs are:​

Answer 1

Step-by-step explanation:

Given  

the number of terms in the expansion of (y^(1/5)+x^(1/10))^55,in which the powers of x and y are free from radical signs are:

• A sign like √ before any number is a radical sign.
• For example √2^2 = (2^2)^1/2 = 2^1 = 2
• We know that Tr + 1 = n Cr
•                                    = 55 C 2 (y^1/5)^55 – r  (x^1/10)^r
•                               = 55 C r y^55 – r / 5 x^r/10
•                                 = 55 C r y ^11 – r/5 x^r/10
• So we have r/5 , r/10 and also 0 <- r < 55
• To get radical free terms, r should be multiple of 10 and (55 – r) should be multiple of 5.
• If r is a multiple of 10
•                           So r = { 0, 10, 20, 30, 40, 50}
• If r is a multiple of 55 – r , then r = {0,5,10,15,20,25,30,35,40,45,50,55}
• Taking common values we get r = {0,10,20, 30,40,50}

 Therefore 6 terms will be radical free.