Answer : (a) 16. 8C4
Sum of the coefficients of the expansion \((\frac{1}{x} +2x)^n\) = 6561
Putting x = 1, (1 + 2)n = 6561 ⇒ 3n = 38
⇒ n = 8
\(\therefore\) Tr + 1 in the expansion of \((\frac{1}{x} +2x)^8\)
= 8Cr \((\frac{1}{x})^{8-r}\) (2x)r = 8Cr 2r x2r-8
Since this term is independent of x, 2r – 8 = 0 ⇒ r = 4.
∴ Reqd. term = 8C4 . 24
= 16 . 8C4.