$$\eqalign{ & {\left( {1 - ax} \right)^{ - 1}}{\left( {1 - bx} \right)^{ - 1}} \cr & = \left( {1 + ax + {a^2}{x^2} + ....} \right)\left( {1 + bx + {b^2}{x^2} + ....} \right) \cr & \therefore \,\,{\text{Co - efficient of }}{x^n} \cr & {x^n} = {b^n} + a{b^{n - 1}} + {a^2}{b^{n - 2}} + .... + {a^{n - 1}}b + {a^n} \cr & \left\{ {{\text{which is a G}}{\text{.P}}{\text{. with }}r = \frac{a}{b}} \right. \cr & \left. {\therefore \,\,{\text{Its sum is}} = \frac{{{b^n}\left[ {1 - {{\left( {\frac{a}{b}} \right)}^{n + 1}}} \right]}}{{1 - \frac{a}{b}}}} \right\} = \frac{{{b^{n + 1}} - {a^{n + 1}}}}{{b - a}} \cr & \therefore \,\,{a_n} = \frac{{{b^{n + 1}} - {a^{n + 1}}}}{{b - a}} \cr} $$

Given series is $$^{20}{C_0} + {\,^{20}}{C_1} + {\,^{20}}{C_2} + ..... + {\,^{28}}{C_8}$$
$$\eqalign{ & = \frac{{\left( {{2^{20}} - {\,^{20}}{C_{10}}} \right)}}{2} - {\,^{20}}{C_9} \cr & = {2^{19}} - \frac{{\left( {^{20}{C_{10}} + 2 \times {\,^{20}}{C_9}} \right)}}{2} \cr} $$

$$\eqalign{ & {\left[ {\left( {{t^{ - 1}} - 1} \right)x + {{\left( {{t^{ - 1}} + 1} \right)}^{ - 1}}{x^{ - 1}}} \right]^8} \cr & = \,{\left[ {\left( {\frac{1}{t} - 1} \right)x + {{\left( {\frac{1}{t} + 1} \right)}^{ - 1}}\frac{1}{x}} \right]^8} \cr} $$
Let $${T_{r + 1}}$$  be the term independent of $$x,$$ then
$$\eqalign{ & {T_{r + 1}} = {\,^8}{C_r}{\left( {\frac{1}{t} - 1} \right)^{8 - r}}{x^{8 - r}} \cdot {\left( {\frac{1}{t} + 1} \right)^{ - r}}{\left( {\frac{1}{x}} \right)^r} \cr & = {\,^8}{C_r}{\left( {\frac{1}{t} - 1} \right)^{8 - r}} \cdot {\left( {\frac{1}{t} + 1} \right)^{ - r}} \cdot {x^{8 - 2r}} \cr & \therefore \,8 - 2r = 0 \cr & \Rightarrow \,r = 4 \cr} $$
$$\therefore {T_5}$$  is the term independent of $$x$$ and
$$\eqalign{ & {T_5} = {\,^8}{C_4}{\left( {\frac{1}{t} - 1} \right)^4} \cdot {\left( {\frac{1}{t} + 1} \right)^{ - 4}} \cr & = {\,^8}{C_4}{\left( {\frac{{1 - t}}{t}} \right)^4} \cdot {\left( {\frac{{1 + t}}{t}} \right)^{ - 4}} \cr & = {\,^8}{C_4}{\left( {\frac{{1 - t}}{{1 + t}}} \right)^4} = \frac{{8 \cdot 7 \cdot 6 \cdot 5}}{{4 \cdot 3 \cdot 2 \cdot 1}}{\left( {\frac{{1 - t}}{{1 + t}}} \right)^4} \cr & = \,70 \cdot {\left( {\frac{{1 - t}}{{1 + t}}} \right)^4} \cr} $$

$$\eqalign{ & \frac{1}{2}{x^2} + \frac{2}{3}{x^3} + \frac{3}{4}{x^4} + \frac{4}{5}{x^5} + ..... \cr & = \left( {1 - \frac{1}{2}} \right){x^2} + \left( {1 - \frac{1}{3}} \right){x^3} + \left( {1 - \frac{1}{4}} \right){x^4} + \left( {1 - \frac{1}{5}} \right){x^5} + ..... \cr & = \left( {{x^2} + {x^3} + {x^4} + {x^5} + .....} \right) + \left( { - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} - \frac{{{x^5}}}{5}.....} \right) \cr & = \left( {x + {x^2} + {x^3} + .....} \right) + \left( { - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} - \frac{{{x^5}}}{5}.....} \right) \cr & = \frac{x}{{1 - x}} + \log \left( {1 - x} \right) \cr} $$

$$\eqalign{ & {\left( {{x^2} - 2 + \frac{1}{{{x^2}}}} \right)^n} = {\left( {x - \frac{1}{x}} \right)^{2n}} \cr & \therefore \,\,{t_{r + 1}} = {\,^{2n}}{C_r}{x^{2n - r}} \cdot {\left( { - \frac{1}{x}} \right)^r} = {\left( { - 1} \right)^r} \cdot {\,^{2n}}{C_r} \cdot {x^{2n - 2r}}. \cr} $$
If $$2n - 2r = 0,$$   the value of the term will not depend on $$x.$$ So, $$r = n.$$
∴ the required number of terms $$ = \left( {2n + 1} \right) - 1 = 2n.$$

$$\eqalign{ & {\left( {1 - 9x + 20{x^2}} \right)^{ - 1}} = {\left[ {\left( {1 - 4x} \right)\left( {1 - 5x} \right)} \right]^{ - 1}} \cr & = \frac{1}{x}\left[ {\frac{{\left( {1 - 4x} \right) - \left( {1 - 5x} \right)}}{{\left( {1 - 4x} \right) \cdot \left( {1 - 5x} \right)}}} \right] = \frac{1}{x}\left[ {{{\left( {1 - 5x} \right)}^{ - 1}} - {{\left( {1 - 4x} \right)}^{ - 1}}} \right] \cr & = \frac{1}{5}\left[ {\left( {5 - 4} \right)x + \left( {{5^2} - {4^2}} \right){x^2} + \left( {{5^3} - {4^3}} \right){x^3} + ..... + \left( {{5^n} - {4^n}} \right){x^n} + .....} \right] \cr & \therefore {\text{coeff}}{\text{. of }}{x^n} = {5^{n + 1}} - {4^{n + 1}} \cr} $$

$$\eqalign{ & \frac{{{e^{7x}} + {e^x}}}{{{e^{3x}}}} = {e^{4x}} + {e^{ - 2x}} \cr & = \left[ {1 + 4x + \frac{{{{\left( {4x} \right)}^2}}}{{2!}} + .....} \right] + \left[ {1 + \left( { - 2x} \right) + \frac{{{{\left( { - 2x} \right)}^2}}}{{2!}} + .....} \right] \cr & \therefore {\text{coeff}}{\text{. of }}{x^n} = \frac{{{4^n}}}{{n!}} + \frac{{{{\left( { - 2} \right)}^n}}}{{n!}} \cr} $$

The given expression is
$${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}$$
We know by binomial theorem, that
$${\left( {x + a} \right)^n} + {\left( {x - a} \right)^n} - 2\left[ {^n{C_0}{x^n} + {\,^n}{C_2}{x^{n - 2}}{a^2} + {\,^n}{C_4}{x^{n - 4}}{a^4} + ......} \right]$$
∴ The given expression is equal to
$$2\left[ {^5{C_0}{x^5} + {\,^5}{C_2}{x^3}\left( {{x^3} - 1} \right) + {\,^5}{C_4}x{{\left( {{x^3} - 1} \right)}^2}} \right]$$
Max. power of $$x$$ involved here is 7, also only +ve integral powers of $$x$$ are involved, therefore given expression is a polynomial of degree 7.

Let $$S = {\,^{20}}{C_0} + {\,^{20}}{C_1} + {\,^{20}}{C_2} + ..... + {\,^{20}}{C_{10}}$$
$$S = {\,^{20}}{C_{20}} + {\,^{20}}{C_{19}} + {\,^{20}}{C_{18}} + ..... + {\,^{20}}{C_{10}}\,\,\,\left( {\because \,{\,^n}{C_r} = {\,^n}{C_{n - r}}} \right)$$
Adding, $$2S = \left( {^{20}{C_0} + {\,^{20}}{C_1} + ..... + {\,^{20}}{C_{20}}} \right) + {\,^{20}}{C_{10}} = {2^{20}} + {\,^{20}}{C_{10}}.$$

We have,
$$\eqalign{ & \sum\limits_{r = 0}^n {r{\,^n}{C_r}\,{x^r}{y^{n - r}}} = \sum\limits_{r = 0}^n {r\frac{n}{r}{\,^{n - 1}}{C_{r - 1}}} {x^{r - 1}}{x^1}{y^{n - r}} \cr & = nx\sum\limits_{r = 0}^n {^{n - 1}{C_{r - 1}}{x^{r - 1}}{y^{\left( {n - 1} \right) - \left( {r - 1} \right)}}} \cr & = nx{\left( {x + y} \right)^{n - 1}} = nx\left[ {\because x + y = 1} \right] \cr} $$