(2 + √x + 4/(√x - 2))^-40

 = \((\frac{x-4+4}{\sqrt x-2})^{-40}\)

 = x^-40(√x - 2)⁴⁰

 = x ^-40(\(\sum\limits_{r=0}^{40}\) ⁴⁰C_r x ^{\(\frac{40-r}2\)}(-2)^r), 0\(\leq\)r \(\leq\) 40

 = \(\sum\limits_{r=0}^{40}\) ⁴⁰C_r x^{\(-(\frac{40-r}2)\)}(-2)²

For independent term-(\(\frac{40+r}2=0\))

⇒ r = -40 which is not positive.

Hence, independent term in expansion of (2 + √x + \(\frac4{\sqrt4 - 2}\))^-40 is 0.

\(\left(2+\sqrt x - \frac 4 {\sqrt x -2}\right)^{-40} = \left(\frac{x- 4 - 4}{\sqrt x - 2}\right)^{-40}\)

\(= (x - 8) ^{-40} ( \sqrt x - 2)^{40}\)

\(= x^{-40} \left(1 - \frac8 x\right)^{-40} (\sqrt x - 2^{40})\)

\(= x^{-40} \left( \displaystyle\sum^{40}_{r=0} \,^{40}C_r \;x^{\frac{40 - r}2} \;(-2)^r\right) \left(1 - \frac8x\right)^{-40}\)

\(= \left(\displaystyle\sum^{40}_{r = 0} (-1)^r 2^r \,^{40}C_r\; x^{\left(\frac{40 + r}2\right)}\right)\left(1 - \frac8x\right)^{-40}\)

\(= \left(\displaystyle\sum^{40}_{r = 0} (-1)^r 2^r \,^{40}C_r\; x^{\left(\frac{40 + r}2\right)}\right)\left(\displaystyle\sum^\infty_{k = 0}\frac{n(n + 1)...(n + k - 1)}{k!} x^k\right), n = 40;k\in Z\)

\(= \displaystyle\sum^{40}_{r = 0} \displaystyle\sum^\infty_{k =0} ( -1)^r 2^r \,^{40}C_r \frac{n (n + 1)....(n + k - 1)}{k!} x^{-\frac{40+r}2+ k} , n = 40;k\in Z\)

For independent term 

\(-\frac{40 + r}{2} + k = 0\)

⇒ \(k = \frac{40 + r}{2}\)

and \(0 \le r\le 40\)

∴ Independent term = \(\displaystyle\sum^{40}_{r = 0\\r = even} 2^r \, ^{40} C_r \frac{40. 41.42. ......\left(\frac{r + 118}2\right)}{\left(\frac{40 + r}{2}\right)!}\)