Binomial Theorem MCQ Questions & Answers in Algebra | Maths

$${t_{r + 1}} = {\,^{4n - 2}}{C_r}{\left( {ix} \right)^r}.$$    The term is real negative if $$r = 2,6,10,.....\,\,.$$
But $$0 \leqslant r \leqslant 4n - 2\,\,{\text{and }}4n - 2 = 2 + \left( {n - 1} \right)4.$$
∴ the required number of terms $$= n.$$

The $$7^{th}$$ term from the end $$= 5^{th}$$  term from beginning
$${T_5} = {\,^{10}}{C_4}{x^6}{\left( { - \frac{2}{{{x^2}}}} \right)^4} = {\,^{10}}{C_4} \cdot {2^4}\left( {\frac{1}{{{x^2}}}} \right)$$

$$\eqalign{ & {\left( {1 - x + {x^2}} \right)^5} = {\left\{ {1 + x\left( {x - 1} \right)} \right\}^5} \cr & {\left( {1 - x + {x^2}} \right)^5} = {\,^5}{C_0} + {\,^5}{C_1}x\left( {x - 1} \right) + {\,^5}{C_2}{x^2}{\left( {x - 1} \right)^2} + {\,^5}{C_3}{x^3}{\left( {x - 1} \right)^3} + ..... \cr} $$
∴ the co-efficient of $${x^3} = - 2 \cdot {\,^5}{C_2} - {\,^5}{C_3}.$$

$$\eqalign{ & {\left[ {{{\left( {x + 4y} \right)}^3}{{\left( {x - 4y} \right)}^3}} \right]^2} = {\left[ {\left\{ {{x^2} - {{\left( {4y} \right)}^2}} \right\}} \right]^6} \cr & = {\left( {{x^2} - 16{y^2}} \right)^6} \cr} $$
$$\therefore $$ No. of terms in the expansion $$ = 7$$

$${T_{r + 1}} = {\left( { - 1} \right)^r}.{\,^n}{C_r}{\left( a \right)^{n - r}}.{\left( b \right)^r}$$      is an expansion of $${\left( {a - b} \right)^n}$$
$$\eqalign{ & \therefore \,\,{5^{th}}\,{\text{term}}\,{\text{ = }}\,{t_5} = {t_{4 + 1}} \cr & = \,{\left( { - 1} \right)^4}{.^n}{C_4}{\left( a \right)^{n - 4}}.{\left( b \right)^4} = {\,^n}{C_4}.{a^{n - 4}}\,.\,{b^4} \cr & {6^{th}}{\text{ term }} = {t_6} = {t_{5 + 1}} \cr & = {\left( { - 1} \right)^5}.{\,^n}{C_5}{\left( a \right)^{n - 5}}{\left( b \right)^5} \cr & {\text{Given }}\,{t_5} + {t_6} = 0 \cr & \therefore \,{\,^n}{C_4}.{a^{n - 4}}.{b^4} + \left( { - {\,^n}{C_5}.{a^{n - 5}}.{b^5}} \right) = 0 \cr & \Rightarrow \,\,\frac{{n!}}{{4!\left( {n - 4} \right)!}}.\frac{{{a^n}}}{{{a^4}}}.{b^4} - \frac{{n!}}{{5!\left( {n - 5} \right)!}}.\frac{{{a^n}{b^5}}}{{{a^5}}} = 0 \cr & \Rightarrow \,\,\frac{{n!.{a^n}{b^4}}}{{4!\left( {n - 5} \right)!.{a^4}}}\left[ {\frac{1}{{\left( {n - 4} \right)}} - \frac{b}{{5.a}}} \right] = 0 \cr & {\text{or, }}\frac{1}{{n - 4}} - \frac{b}{{5a}} = 0 \cr & \Rightarrow \,\,\frac{a}{b} = \frac{{n - 4}}{5} \cr} $$

$$\eqalign{ & {8^{2n}} - {\left( {62} \right)^{2n + 1}} \cr & = {\left( {64} \right)^n} - {\left( {62} \right)^{2n + 1}} = {\left( {63 + 1} \right)^n} - {\left( {63 - 1} \right)^{2n + 1}} \cr & = \left[ {^n{C_0}{{\left( {63} \right)}^n} + {\,^n}{C_1}{{\left( {63} \right)}^{n - 1}} + {\,^n}{C_2}{{\left( {63} \right)}^{n - 2}} + ..... + {\,^n}{C_{n - 1}}\left( {63} \right) + {\,^n}{C_n}} \right] \cr & = \left[ {^{2n + 1}{C_0}{{\left( {63} \right)}^{2n + 1}} - {\,^{2n + 1}}{C_1}{{\left( {63} \right)}^{2n}} + {\,^{2n + 1}}{C_2}{{\left( {63} \right)}^{2n - 1}} - ..... + \,{{\left( { - 1} \right)}^{2n + 1}}{\,^{2n + 1}}{C_{2 + 1}}} \right] \cr & = 63 \times \left[ {^n{C_0}{{\left( {63} \right)}^{n - 1}} + {\,^n}{C_1}{{\left( {63} \right)}^{n - 2}} + {\,^n}{C_2}{{\left( {63} \right)}^{n - 3}} + .....} \right] + 1 - 63 \times \left[ {^{2n + 1}{C_0}{{\left( {63} \right)}^{2n}} - {\,^{2n + 1}}{C_1}{{\left( {63} \right)}^{2n - 1}} + .....} \right] + 1 \cr & \Rightarrow \,\,63 \times {\text{some integral value}} + 2 \cr} $$
$$ \Rightarrow \,\,{8^{2n}} - {\left( {62} \right)^{2n + 1}}$$    when divided by 9 leaves 2 as the remainder.

We have, $${T_2} = 14\,{a^{\frac{5}{2}}}$$
$$\eqalign{ & \Rightarrow {\,^n}{C_1}{\left( {{a^{\frac{1}{{13}}}}} \right)^{n - 1}}\left( {{a^{\frac{3}{2}}}} \right) = 14{a^{\frac{5}{2}}} \cr & \Rightarrow \,n{a^{\frac{{n - 1}}{{13}} + \frac{3}{2}}} = 14{a^{\frac{5}{2}}} \cr & \Rightarrow \,n = 14 \cr & \Rightarrow \,\frac{{{\,^n}{C_3}}}{{{\,^n}{C_2}}} = \frac{{{\,^{14}}{C_3}}}{{{\,^{14}}{C_2}}} = \frac{{12}}{3} = 4 \cr} $$

The last term $$ = {\,^n}{C_n} \cdot {\left( { - \frac{1}{{\sqrt 2 }}} \right)^n} = {\left( {\frac{1}{{3 \cdot \root 3 \of 9 }}} \right)^{{{\log }_3}8}}\left( {{\text{from the question}}} \right)$$
$$\eqalign{ & \therefore \,\,{\left( { - 1} \right)^n} \cdot {\left( {\frac{1}{2}} \right)^{\frac{n}{2}}} = {\left( {\frac{1}{{{3^{\frac{5}{3}}}}}} \right)^{{{\log }_3}8}} \cr & = {3^{ - \frac{5}{3} \cdot 3{{\log }_3}2}} = {3^{{{\log }_3}{2^{ - 5}}}} = {2^{ - 5}} = {\left( {\frac{1}{2}} \right)^5}. \cr & \therefore \,\,n = 10. \cr & {\text{So, }}{t_5} = {\,^{10}}{C_4} \cdot {\left( {{2^{\frac{1}{3}}}} \right)^6} \cdot {\left( { - \frac{1}{{\sqrt 2 }}} \right)^4} = {\,^{10}}{C_4} = {\,^{10}}{C_{10 - 4}}. \cr} $$

We have, $${\left( {1 + x + {x^2} + {x^3} + {x^4}} \right)^n}{\left( {1 - x} \right)^{n + 3}}$$
$$\eqalign{ & = \frac{{{{\left( {{x^5} - 1} \right)}^n}}}{{{{\left( {1 - x} \right)}^n}}} \cdot {\left( {x - 1} \right)^{n + 3}} = {\left( {{x^5} - 1} \right)^n}{\left( {x - 1} \right)^3} \cr & = \left( { + {x^3} - 3{x^2} + 3x - 1} \right)\sum\limits_{r = 0}^n {^n{C_r}{{\left( { - 1} \right)}^r}{x^{5r}}} \cr & = + \sum\limits_{r = 0}^n {^n{C_r}{{\left( { - 1} \right)}^r}{x^{5r + 3}} + 3\sum\limits_{r = 0}^n {^n{C_r}{{\left( { - 1} \right)}^r}{x^{5r + 2}}} } - 3\sum\limits_{r = 0}^n {^n{C_r}} {\left( { - 1} \right)^r}{x^{5r + 1}} + 3\sum\limits_{r = 0}^n {^n{C_r}} {\left( { - 1} \right)^r}{x^{5r}} \cr} $$
For term containing $$x^{83},$$ we have $$5r + 3 = 83$$
$$ \Rightarrow $$ $$r = 16$$  whereas $$5r + 2 = 83, 5r + 1 = 83$$     and $$5r = 83$$  give no integral value of $$r.$$ Hence, there is only one term containing $$x^{83}$$ whose coefficient
$$\eqalign{ & = - {\,^n}{C_{16}} = - {\,^n}{C_{2\lambda }} \cr & \therefore 2\lambda = 16 \cr & \Rightarrow \lambda = 8. \cr} $$

$$\eqalign{ & {\left( {1 + {t^2}} \right)^{12}}\left( {1 + {t^{12}}} \right)\left( {1 + {t^{24}}} \right) \cr & = \left( {1 + {t^{12}} + {t^{24}} + {t^{36}}} \right){\left( {1 + {t^2}} \right)^{12}} \cr} $$
∴ Co-eff. of $${t^{24}} = 1 \times $$  Co-eff. of $${t^{24}}$$ in $${\left( {1 + {t^2}} \right)^{12}} + 1 \times $$   Co-eff. of $${t^{12}}$$ in $${\left( {1 + {t^2}} \right)^{12}} + 1 \times $$   constant term in $${\left( {1 + {t^2}} \right)^{12}}$$
$$\eqalign{ & = {\,^{12}}{C_{12}} + {\,^{12}}{C_6} + {\,^{12}}{C_0} \cr & = 1 + {\,^{12}}{C_6} + 1 \cr & = {\,^{12}}{C_6} + 2 \cr} $$