(3)

We have, ${(\frac{(x+1)}{(x^{\frac{2}{3}}+1–x^{\frac{1}{3}})}–\frac{(x–1)}{(x–x^{\frac{1}{2}})})}^{10}$

$=\left(\right(x^{\frac{1}{3}}+1)–{\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right))}^{10}={(x^{\frac{1}{3}}–\frac{1}{\sqrt{x}})}^{10}$

$T_{r+1}={}^{10}C_r{\left(x\right)}^{\frac{10–r}{3}}{(–1)}^r{\left(x\right)}^{–\frac{r}{2}}$

Since, the term is independent of x, then $\frac{10–r}{3}–\frac{r}{2}=0$

$\Rightarrow$  (20 – 2r) – 3r = 0 $\Rightarrow$ r = 4.

Hence, the required term is ${}^{10}C_4{(–1)}^4=210$.

(4)

Given, ${(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}})}^n, n\in N$

Now, $T_{r+1}={}^{\mathbf{n}}\mathit{C}_r\mathbf{·}{\mathbf{\left(}\sqrt[\mathbf{3}]{\mathbf{2}}\mathbf{\right)}}^{\mathit{n}\mathbf{–}\mathit{r}}{\mathbf{\left(}\frac{\mathbf{1}}{\sqrt[\mathbf{3}]{\mathbf{3}}}\mathbf{\right)}}^r$

$\therefore T_{15}={}^{\mathbf{n}}\mathit{C}_{\mathbf{14}}\mathbf{·}{\mathbf{\left(}\sqrt[\mathbf{3}]{\mathbf{2}}\mathbf{\right)}}^{\mathit{n}\mathbf{–}\mathbf{14}}{\mathbf{\left(}\frac{\mathbf{1}}{\sqrt[\mathbf{3}]{\mathbf{3}}}\mathbf{\right)}}^{\mathbf{14}}$

${15}^{\mathrm{th}}$ term from end = $T_{n–13}$ from beginning

$\Rightarrow T_{n–13}={}^{\mathrm{n}}C_{\mathrm{n}–14}·{\left(\sqrt[3]{2}\right)}^{14}{\left(\frac{1}{\sqrt[3]{3}}\right)}^{\mathrm{n}–14}$

But $\frac{T_{15}}{T_{n–13}}=\frac{{}^{n}C_{14}{\left(\sqrt[3]{2}\right)}^{n–14}{\left({\displaystyle \frac{1}{\sqrt[3]{3}}}\right)}^{14}}{{}^{n}C_{n–14}{\left(\sqrt[3]{2}\right)}^{14}{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n–14}}=\frac{1}{6}$

$\Rightarrow {\left(\sqrt[3]{2}\right)}^{n–28}\times {\left(\sqrt[3]{3}\right)}^{n–28}=6^{–1} \Rightarrow {\left(\sqrt[3]{6}\right)}^{n–28}=6^{–1}$

$\Rightarrow \frac{n–28}{3}=–1 \Rightarrow n=25$

So, ${}^{n}C_3={}^{25}C_3=2300$.

(4)

$T_{r+1}={}^{1016}C_{r }{5}^{\frac{1016–r}{2}}{7}^{r/8}$, represents the ${(r+1)}^{\mathrm{th}}$ term of ${(5^{1/2}+7^{1/8})}^{1016}$.

As $T_{r+1}$ represent an integral term when r is a multiple of 8

i.e., r = 0, 8, 16, 24, ....., 1016

Now, 1016 = 0 + (n – 1)8          [As this form an A.P.]

$\Rightarrow n–1=127 \Rightarrow n=128$

So, 128 terms are integral terms.

(1)

Coefficient of $5^{\mathrm{th}}, 6^{\mathrm{th}}$ and $7^{\mathrm{th}}$ terms of the expansion of ${(1+x)}^{n+4}$ are ${}^{n+4}C_4, {}^{n+4}C_5$ and ${}^{n+4}C_6$ respectively. As ${}^{n+4}C_4, {}^{n+4}C_5$ and ${}^{n+4}C_6$ are in A.P.

$\Rightarrow 2\times {}^{n+4}C_5={}^{n+4}C_4+{}^{n+4}C_6$

$\Rightarrow \frac{2}{5(n–1)}=\frac{1}{n(n–1)}+\frac{1}{30} \Rightarrow \frac{2}{5(n–1)}=\frac{30+n^2–n}{30n(n–1)}$

$\Rightarrow 12n=30+n^2–n \Rightarrow n^2–13n+30=0$

$\Rightarrow (n–3)(n–10)=0 \Rightarrow n=3$          [$\because n\ne 10$]

Here, n + 4 = 3 + 4 = 7

$\therefore$   Largest binomial coefficient in expansion ${(1+x)}^7$ = Coefficient of middle term = ${}^{7}C_4 \mathrm{or} {}^{7}C_3=35$.

(2)

General term = ${}^{n}C_r{\left(7^{1/3}\right)}^{n–r}{\left({11}^{1/12}\right)}^r$

                      = ${}^{n}C_r{\left(7\right)}^{\frac{n–r}{3}}{\left(11\right)}^{r/12}$

For integral terms, r must be multiple of 12

$\therefore$   r = 12k, k = 0, 1, 2, .....

Total number of values of r = 183

Hence, max r = 12(182) = 2184

Minimum value of n = 2184.