Given expansion is (y1/5+x1/10)55 The general term is Tr+1=55Cr(y1/5)55−r⋅(x101)r Tr+1 would free from radical sign if powers of y and x are integers. i.e. 555−r and 10r are integer. ⇒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=10k−10⇒k=6 Thus, the six terms of the given expansion in which x and y are free from radical signs.