The term independent of x in the expansion of (x2/3−x1/3+1x+1−x−x1/2x−1)10, x=1, is equal to
Answer: 210
Solutions · 1
Official solution
As in Question 31: this simplifies to (x1/3−x−1/2)10.
\[T_{r+1}=\binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r\]
Independent of x: 310−r−2r=0⇒20−2r−3r=0⇒r=4
\[T_5=\binom{10}{4}=210\]