Binomial Theorem MCQ Questions & Answers in Algebra | Maths

$${\text{We have }}{2^n} = 4096 = {2^{12}}$$
$$ \Rightarrow \,\,n = 12;$$   the greatest co-eff.
= co-eff. of middle term. So middle term
$$ = {t_7};{t_7} = {t_{6 + 1}}$$
$$ \Rightarrow \,\,{\text{co-eff}}{\text{. of }}{t_7} = {\,^{12}}{C_6} = \frac{{12!}}{{6!6!}} = 924.$$

$$\eqalign{ & {\left( {\sqrt 2 + 1} \right)^6} = P + g = {\,^6}{C_0}{\left( {\sqrt 2 } \right)^6} + {\,^6}{C_1}{\left( {\sqrt 2 } \right)^5} + ..... + {\,^6}{C_6},\,{\text{where }}P = {\text{integer, }}0 < g < 1. \cr & {\left( {\sqrt 2 - 1} \right)^6} = f = {\,^6}{C_0}{\left( {\sqrt 2 } \right)^6} - {\,^6}{C_1}{\left( {\sqrt 2 } \right)^5} + .....,0 < f < 1 \cr & \therefore \,\,P + f + g = 2\left\{ {^6{C_0} \cdot {2^3} + {\,^6}{C_2} \cdot {2^2} + {\,^6}{C_4} \cdot 2 + {\,^6}{C_6}} \right\} \cr & P + f + g = 2\left( {8 + 15 \times 4 + 15 \times 2 + 1} \right) = 198 \cr & \therefore \,\,197 < P + g < 198 \cr & \therefore \,\,197 < {\left( {\sqrt 2 + 1} \right)^6} < 198 \cr & \therefore \,\,n < {\left( {\sqrt 2 + 1} \right)^6} \cr} $$
⇒ the greatest value of the natural number $$n = 197.$$

Expression $$ = \frac{1}{{n!}}\left\{ {^n{C_1} + {\,^n}{C_3} + {\,^n}{C_5} + .....} \right\} = \frac{1}{{n!}} \cdot {2^{n - 1}}\,{\text{for all }}n \in N.$$

We have, $$\left( {1 - x} \right)\left( {{B_0} + {B_1}x + {B_2}{x^2} + ..... + {B_{n - 1}}{x^{n - 1}} + {B_n}{x^n} + .....} \right)$$
$$ = {e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ..... + \frac{{{x^n}}}{{n!}} + .....\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right)$$
Hence, equating the coefficients of $$x^n$$ on both sides of (1)
$${\text{we get, }}{B_n} - {B_{n - 1}} = \frac{1}{{n!}}.$$

Expression $$ = \frac{1}{{{x^{10}}}}\left( {1 - 2x + {x^2}} \right){\left( {1 + {x^2}} \right)^{10}}$$
$$ = \frac{1}{{{x^{10}}}}{\left( {1 + {x^2}} \right)^{10}} - \frac{2}{{{x^9}}}{\left( {1 + {x^2}} \right)^{10}} + \frac{1}{{{x^8}}}{\left( {1 + {x^2}} \right)^{10}}.$$
∴ the term independent of $$x = {\,^{10}}{C_5} - 2 \times 0 + {\,^{10}}{C_4}.$$

$$\eqalign{ & {\left( {1 + \sqrt 2 + \root 3 \of 3 } \right)^6} = {\,^6}{C_0} + {\,^6}{C_1}\left( {\sqrt 2 + \root 3 \of 3 } \right) + {\,^6}{C_2}{\left( {\sqrt 2 + \root 3 \of 3 } \right)^2} + ..... + {\,^6}{C_6}{\left( {\sqrt 2 + \root 3 \of 3 } \right)^6} \cr & {\text{In }}{\left( {\sqrt 2 + \root 3 \of 3 } \right)^6},{t_{r + 1}} = {\,^6}{C_r} \cdot {\left( {\sqrt 2 } \right)^{6 - r}} \cdot {\left( {\root 3 \of 3 } \right)^r}. \cr} $$
It is rational if $$r = 0, 6.$$   Similarly in other cases.
∴ the number of rational terms $$= 1 + 0 + 1 + 1 + 1 + 1 + 2 = 7.$$

$$\eqalign{ & {3^{2n}} = {\left( {1 + 8} \right)^n} = {\,^n}{C_0} + {\,^n}{C_1} \cdot 8 + {\,^n}{C_2} \cdot {8^2} + ..... + {\,^n}{C_n}{8^n} \cr & \therefore \,\,\frac{{{3^{2n}}}}{8} = \frac{1}{8} + \left( {^n{C_1} + {\,^n}{C_2} \cdot 8 + ..... + {\,^n}{C_n} \cdot {8^7}} \right) = \frac{1}{8} + {\text{integer}}{\text{.}} \cr} $$

$$\eqalign{ & {\left( {1 - 2\sqrt x } \right)^{50}} = {\,^{50}}{C_0} - {\,^{50}}{C_1}\,2\sqrt x + {\,^{50}}{C_2}{\left( {2\sqrt x } \right)^2}\,\,\,\,\,.....\left( 1 \right) \cr & {\left( {1 + 2\sqrt x } \right)^{50}} = {\,^{50}}{C_0} + {\,^{50}}{C_1}\,2\sqrt x - {\,^{50}}{C_2}{\left( {2\sqrt x } \right)^2} + ..... + {\,^{50}}{C_3}{\left( {2\sqrt x } \right)^3} - {\,^{50}}{C_4}{\left( {2\sqrt x } \right)^4}\,\,\,\,.....\left( 2 \right) \cr} $$
Adding equation (1) and (2)
$$\eqalign{ & {\left( {1 - 2\sqrt x } \right)^{50}} + {\left( {1 + 2\sqrt x } \right)^{50}} \cr & = 2\left[ {^{50}{C_0} + {\,^{50}}{C_2}{2^2}x + {\,^{50}}{C_4}{2^3}{x^2} + .....} \right] \cr} $$
Putting $$x = 1,$$  we get above as $$\frac{{{3^{50}} + 1}}{2}$$

$${t_{r + 1}} = {\,^{10}}{C_r} \cdot {2^{\frac{{10 - r}}{2}}} \cdot {3^{\frac{r}{5}}}.$$     This is rational if $$\frac{r}{5}$$ and $$5 - \frac{r}{2}$$  are integers.
Also $$0 \leqslant r \leqslant 10.$$
Now, $$r$$ must be a multiple of 10(LCM of 2 and 5). So $$r = 0, 10.$$
∴ the rational terms are $$^{10}{C_0} \cdot {2^5} \cdot {3^0},{\,^{10}}{C_{10}} \cdot {2^0} \cdot {3^2}.$$

Given, $$T_4 = 200$$
$$\eqalign{ & \Rightarrow {\,^6}{C_3}{\left( {\sqrt {{x^{\left( {\frac{1}{{\log x + 1}}} \right)}}} } \right)^3}{\left( {{x^{\frac{1}{{12}}}}} \right)^3} = 200 \cr & \Rightarrow 20 \cdot {x^{\frac{3}{{2\left( {\log x + 1} \right)}} + \frac{1}{4}}} = 200 \cr & \Rightarrow {x^{\left\{ {\frac{3}{{2\left( {\log x + 1} \right)}} + \frac{1}{4}} \right\}}} = 10 \cr & \Rightarrow \frac{3}{{2\left( {\log x + 1} \right)}} + \frac{1}{4} = {\log _x}10 = \frac{1}{{{{\log }_{10}}x}} \cr & \Rightarrow \frac{3}{{2\left( {y + 1} \right)}} + \frac{1}{4} = \frac{1}{y}{\text{ where }}y = {\log _{10}}x \cr & \Rightarrow y = - 4{\text{ or }}y = 1 \cr & \Rightarrow {\log _{10}}x = - 4{\text{ or }}{\log _{10}}x = 1 \cr & \Rightarrow x = {10^{ - 4}}{\text{ or 10}} \cr & \Rightarrow x = 10\left( {\because x > 1} \right) \cr} $$