Binomial Theorem MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & \left[ {{{\left( {\frac{{1 + \sqrt {3x + 1} }}{2}} \right)}^7} - {{\left( {\frac{{1 - \sqrt {3x + 1} }}{2}} \right)}^7}} \right] \cr & = \frac{1}{{\sqrt {3x + 1} }}\left[ {\frac{{{{\left( {1 + \sqrt {3x + 1} } \right)}^7} - {{\left( {1 - \sqrt {3x + 1} } \right)}^7}}}{{{2^7}}}} \right] \cr & = \frac{1}{{\sqrt {3x + 1} }}\left[ {\frac{{2\left\{ {^7{C_1}\left( {\sqrt {3x + 1} } \right) + {\,^7}{C_3}{{\left( {\sqrt {3x + 1} } \right)}^3} + {\,^7}{C_5}{{\left( {\sqrt {3x + 1} } \right)}^5} + {\,^7}{C_7}{{\left( {\sqrt {3x + 1} } \right)}^7}} \right\}}}{{{2^7}}}} \right] \cr & = \frac{1}{{{2^6}}}\left[ {^7{C_1} + {\,^7}{C_3}\left( {3x + 1} \right) + {\,^7}{C_5}{{\left( {3x + 1} \right)}^2} + {\,^7}{C_7}{{\left( {3x + 1} \right)}^3}} \right] \cr} $$
Clearly above is a polynomial of degree 3 in $$x.$$

Put $${\log _{10}}x = y,$$   the given expression becomes $${\left( {x + {x^y}} \right)^5}.$$
$$\eqalign{ & {T_3} = {\,^5}{C_2} \cdot {x^3}{\left( {{x^y}} \right)^2} = 10{x^{3 + 2y}} = {10^6}\left( {{\text{given}}} \right) \cr & \Rightarrow \left( {3 + 2y} \right){\log _{10}}x = 5{\log _{10}}10 = 5 \cr & \Rightarrow \left( {3 + 2y} \right)y = 5 \cr & \Rightarrow y = 1, - \frac{5}{2} \cr & \Rightarrow {\log _{10}}x = 1{\text{ or }}{\log _{10}}x = - \frac{5}{2} \cr & \therefore x = 10{\text{ or }}x = {\left( {10} \right)^{ - \frac{5}{2}}} \cr} $$

No. of terms in the expansion $$ = {\,^{n + 3 - 1}}{C_{3 - 1}}$$
$$\eqalign{ & \therefore {\,^{n + 2}}{C_2} = 45 \cr & \Rightarrow n = 8 \cr} $$

$$\eqalign{ & \left( {^7{C_0} + {\,^7}{C_1}} \right) + \left( {^7{C_1} + {\,^7}{C_2}} \right) + ..... + \left( {^7{C_6} + {\,^7}{C_7}} \right) \cr & = {\,^8}{C_1} + {\,^8}{C_2} + ..... + {\,^8}{C_7} = {\,^8}{C_0} + {\,^8}{C_1} + {\,^8}{C_2} + ..... + {\,^8}{C_7} + {\,^8}{C_8} - \left( {^8{C_0} + {\,^8}{C_8}} \right) \cr & = {2^8} - 1\left( {1 + 1} \right) = {2^8} - 2. \cr} $$

We know that, $${\left( {1 + x} \right)^{20}} = {\,^{20}}{C_0} + {\,^{20}}{C_1}x + {\,^{20}}{C_2}{x^2} + .....{\,^{20}}{C_{10}}{x^{10}} + ..... + {\,^{20}}{C_{20}}{x^{20}}$$
$$\eqalign{ & {\text{Put }}x = - 1,\left( 0 \right) = {\,^{20}}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... + {\,^{20}}{C_{10}} - {\,^{20}}{C_{11}}..... + {\,^{20}}{C_{20}} \cr & \Rightarrow 0 = 2\left[ {^{20}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... - {\,^{20}}{C_9}} \right] + {\,^{20}}{C_{10}} \cr & \Rightarrow {\,^{20}}{C_{10}} = 2\left[ {^{20}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... - {\,^{20}}{C_9} + {\,^{20}}{C_{10}}} \right] \cr & \Rightarrow {\,^{20}}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... + {\,^{20}}{C_{10}} = \frac{1}{2}{\,^{20}}{C_{10}} \cr} $$

Expression $$ = \frac{{{{\left( {1 + x} \right)}^6}\left\{ {1 - {{\left( {1 + x} \right)}^{10}}} \right\}}}{{1 - \left( {1 + x} \right)}} = \frac{{{{\left( {1 + x} \right)}^{16}} - {{\left( {1 + x} \right)}^6}}}{x}$$
∴ the required co-efficient
= the co-efficient of $$x^7$$ in $$\left\{ {{{\left( {1 + x} \right)}^{16}} - {{\left( {1 + x} \right)}^6}} \right\}$$
$$ = {\,^{16}}{C_7} = {\,^{16}}{C_{16 - 7}}.$$

Given, $$\frac{{{T_7}}}{{{T_{n - 7 + 2}}}} = \frac{1}{6}$$
$$\eqalign{ & \Rightarrow \frac{{^n{C_6}{{\left( {\root 3 \of 2 } \right)}^{n - 6}}{{\left( {\frac{1}{{\root 3 \of 3 }}} \right)}^6}}}{{^n{C_{n - 6}}{{\left( {\root 3 \of 2 } \right)}^6}{{\left( {\frac{1}{{\root 3 \of 3 }}} \right)}^{n - 6}}}} = \frac{1}{6} \cr & \Rightarrow {2^{\frac{{n - 12}}{3}}} \cdot {3^{\frac{{n - 12}}{3}}} = \frac{1}{6} \cr & \Rightarrow {6^{\frac{{n - 12}}{3}}} = {6^{ - 1}} \cr & \therefore \frac{{n - 12}}{3} = - 1 \cr & \Rightarrow n = 9 \cr} $$

$$\eqalign{ & \sum\limits_{r = 0}^n {^n{C_r}\sin \left( {rx} \right)} = \operatorname{Im} \left( {\sum\limits_{r = 0}^n {^n{C_r}{e^{irx}}} } \right) \cr & = \operatorname{Im} \left( {\sum\limits_{r = 0}^n {^n{C_r}{{\left( {{e^{ix}}} \right)}^r}} } \right) = \operatorname{Im} \left( {{{\left( {1 + {e^{ix}}} \right)}^n}} \right) \cr & = \operatorname{Im} {\left( {1 + \cos x + i\sin x} \right)^n} \cr & = \operatorname{Im} {\left( {2{{\cos }^2}\frac{x}{2} + 2i\sin \frac{x}{2} \cdot \cos \frac{x}{2}} \right)^n} \cr & = \operatorname{Im} {\left( {2\cos \frac{x}{2}\left( {\cos \frac{x}{2} + i\sin \frac{x}{2}} \right)} \right)^n} \cr & = {2^n} \cdot {\cos ^n}\frac{x}{2} \cdot \sin \frac{{nx}}{2} \cr} $$

The general term $${T_r} = \left( {r - 1} \right){\,^{15}}{C_r},r = 2,3,4,.....,15$$
$$\eqalign{ & \therefore {T_r} = r{\,^{15}}{C_r} - {\,^{15}}{C_r} \cr & = 15 \cdot {\,^{14}}{C_{r - 1}} - {\,^{15}}{C_r}\left[ {\because r \cdot {\,^n}{C_r} = n \cdot {\,^{n - 1}}{C_{r - 1}}} \right] \cr & \therefore \sum\limits_{r = 2}^{15} {{T_r} = 15\left[ {^{14}{C_1} + {\,^{14}}{C_2} + ..... + {\,^{14}}{C_{14}}} \right]} - \left[ {^{15}{C_2} + {\,^{15}}{C_3} + ..... + {\,^{15}}{C_{15}}} \right] \cr & = 15\left[ {{2^{14}} - 1} \right] - \left[ {{2^{15}} - 1 - 15} \right] \cr & = \left( {15 - 2} \right){2^{14}} + 1 = 13 \cdot {2^{14}} + 1 \cr} $$

Expression $$ = {\left( {1 + {x^2}} \right)^{40}} \cdot {\left( {x + \frac{1}{x}} \right)^{ - 10}} = {\left( {1 + {x^2}} \right)^{30}} \cdot {x^{10}}.$$
The co-efficient of $$x^{20}$$ in $${x^{10}}{\left( {1 + {x^2}} \right)^{30}}$$
= the co-efficient of $$x^{10}$$ in $${\left( {1 + {x^2}} \right)^{30}}$$
$$ = {\,^{30}}{C_5} = {\,^{30}}{C_{30 - 5}}.$$