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.
The number of terms in the expansion of (y^1 / 5+x^1 / 10 )^55, in which powers of x and y are free from radical signs are
JEE Main 2012 — Mathematics Algebra
The number of terms in the expansion of (y1/5+x1/10)55, in which powers of x and y are free from radical signs are
Held on 12 May 2012 · Verified 6 Jul 2026.
Options
- A
six
- B
twelve
- C
seven
- D
five
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