The term independent of x in the expansion of ((x+1)(x2/3+1–x1/3)–(x–1)(x–x1/2))10, x>1, is: [2025]
(3)
We have, ((x+1)(x23+1–x13)–(x–1)(x–x12))10
=((x13+1)–(x+1x))10=(x13–1x)10
Tr+1=Cr10(x)10–r3(–1)r(x)–r2
Since, the term is independent of x, then 10–r3–r2=0
⇒ (20 – 2r) – 3r = 0 ⇒ r = 4.
Hence, the required term is C410(–1)4=210.