Binomial Theorem MCQ Questions & Answers in Algebra | Maths

Consider $$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$$
$$\eqalign{ & = \left( {1 + ax + b{x^2}} \right)\left[ {^{18}{C_0} - {\,^{18}}{C_1}\left( {2x} \right) + {\,^{18}}{C_2}{{\left( {2x} \right)}^2} - {\,^{18}}{C_3}{{\left( {2x} \right)}^3} + {\,^{18}}{C_4}{{\left( {2x} \right)}^4} - ......} \right] \cr & {\text{Co - eff}}{\text{. of }}{x^3} = {\,^{18}}{C_3}{\left( { - 2} \right)^3} + a.{\left( { - 2} \right)^2}.{\,^{18}}{C_2} + b\left( { - 2} \right).{\,^{18}}{C_1} = 0 \cr & {\text{Co - eff}}{\text{. of }}{x^3} = - {\,^{18}}{C_3}.8 + a \times 4.{\,^{18}}{C_2} - 2b \times 18 = 0 \cr & = - \frac{{18 \times 17 \times 16}}{6}.8 + \frac{{4a + 18 \times 17}}{2} - 36\,b = 0 \cr & = - 51 \times 16 \times 8 + a \times 36 \times 17 - 36b = 0 \cr & = - 34 \times 16 + 51a - 3b = 0 \cr & = 51a - 3b = 34 \times 16 = 544 \cr & = 51a - 3b = 544\,\,\,\,\,......\left( {\text{i}} \right) \cr} $$
Only option number (B) satisfies the equation number (i)

$$\eqalign{ & \because \,\,{T_4} = 20 \times {8^7} \cr & \Rightarrow \,{\,^6}{C_3}{\left( {\frac{2}{x}} \right)^3} \times {\left( {{x^{{{\log }_8}x}}} \right)^3} = 20 \times {8^7} \cr & \Rightarrow \,\,8 \times 20 \times {\left( {\frac{{{x^{{{\log }_8}x}}}}{x}} \right)^3} = 20 \times {8^7} \cr & \Rightarrow \,\,\frac{{{x^{{{\log }_8}x}}}}{x} = 64 \cr} $$
Now, take $${{{\log }_8}}$$ on both sides, we get
$$\eqalign{ & {\left( {{{\log }_8}x} \right)^2} - \left( {{{\log }_8}x} \right) = 2 \cr & \Rightarrow \,\,{\log _8}x = - 1\,\,\,\,\,\,{\text{or, }}{\log _8}x = 2 \cr & \Rightarrow \,\,x = \frac{1}{8}\,\,\,\,\,\,\,{\text{or, }}x = {8^2} \cr} $$

The sum of all the numerical co-efficients in the expansion is obtained by putting $$x = 1, y = 1.$$

$$\eqalign{ & \left( {x + {\,^n}{C_0}} \right)\left( {x + 3 \cdot {\,^n}{C_1}} \right)\left( {x + 5 \cdot {\,^n}{C_2}} \right).....\left( {x + \left( {2n + 1} \right) \cdot {\,^n}{C_n}} \right) \cr & = \,{x^{n + 1}} + {x^n}\left\{ {^n{C_0} + 3 \cdot {\,^n}{C_1} + 5 \cdot {\,^n}{C_2} + ..... + \left( {2n + 1} \right) \cdot {\,^n}{C_n}} \right\} + ..... \cr & {\text{Coeff}} \cdot \,{\text{of}}\,{x^n} \cr & = {\,^n}{C_0} + 3 \cdot {\,^n}{C_1} + 5 \cdot {\,^n}{C_2} + ..... + \left( {2n + 1} \right) \cdot {\,^n}{C_n} \cr & = \,1 + \left( {^n{C_1} + 2 \cdot {\,^n}{C_1}} \right) + \left( {^n{C_2} + 4 \cdot {\,^n}{C_2}} \right) + ..... + \left( {^n{C_n} + 2n \cdot {\,^n}{C_n}} \right) \cr & = \,\left( {1 + {\,^n}{C_1} + ..... + {\,^n}{C_n}} \right) + 2\left( {^n{C_1} + 2 \cdot {\,^n}{C_2} + ..... + n \cdot {\,^n}{C_n}} \right) \cr & = \,{2^n} + 2\left[ {n + 2 \cdot \frac{{n\left( {n - 1} \right)}}{{2!}} + 3 \cdot \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}} + ..... + n \cdot 1} \right] \cr & = \,{2^n} + 2n\left[ {1 + {\,^{n - 1}}{C_1} + {\,^{n - 1}}{C_2} + ..... + {\,^{n - 1}}{C_{n - 1}}} \right] \cr & = \,{2^n} + 2n \cdot {2^{n - 1}} = {2^n}\left( {1 + n} \right) = \left( {n + 1} \right) \cdot 2^n \cr} $$

Let $$e^x = z,$$  then
$$\eqalign{ & {e^{{e^x}}} = {e^z} = \sum\limits_{k = 0}^\infty {\frac{{{z^k}}}{{k!}}} = \sum\limits_{k = 0}^\infty {\frac{{{{\left( {{e^x}} \right)}^k}}}{{k!}}} = \sum\limits_{k = 0}^\infty {\frac{{{e^{kx}}}}{{k!}}} \cr & = \left( {1 + \frac{{{e^x}}}{{1!}} + \frac{{{e^{2x}}}}{{2!}} + \frac{{{e^{3x}}}}{{3!}} + .....\,{\text{to }}\infty } \right) \cr & = 1 + \frac{1}{{1!}}\left( {\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} } \right) + \frac{1}{{2!}}\left( {\sum\limits_{n = 0}^\infty {\frac{{{{\left( {2x} \right)}^n}}}{{n!}}} } \right) + \frac{1}{{3!}}\left( {\sum\limits_{n = 0}^\infty {\frac{{{{\left( {3x} \right)}^n}}}{{n!}}} } \right) + {\text{to}}.....\infty \cr & \therefore {\text{Coefficient of }}{x^n}{\text{ in }}{e^{{e^x}}} \cr & = \frac{1}{{1!}}\left( {\frac{1}{{n!}}} \right) + \frac{1}{{2!}}\left( {\frac{{{2^n}}}{{n!}}} \right) + \frac{1}{{3!}}\left( {\frac{{{3^n}}}{{n!}}} \right) + .....\,{\text{to }}\infty \cr & = \frac{1}{{n!}}\left( {\frac{1}{{1!}} + \frac{{{2^n}}}{{2!}} + \frac{{{3^n}}}{{3!}} + .....\,{\text{to }}\infty } \right) \cr} $$

Expression = co-efficient of $$x^3$$ in $$\left\{ {{{\left( {1 + x} \right)}^{10}} + {{\left( {1 + x} \right)}^{11}} + {{\left( {1 + x} \right)}^{12}} + ..... + {{\left( {1 + x} \right)}^{20}}} \right\}$$
= co-efficient of $$x^3$$ in $$\frac{{{{\left( {1 + x} \right)}^{10}}\left\{ {1 - {{\left( {1 + x} \right)}^{11}}} \right\}}}{{1 - \left( {1 + x} \right)}}$$
= co-efficient of $$x^4$$ in $$\left\{ {{{\left( {1 + x} \right)}^{21}} - {{\left( {1 + x} \right)}^{10}}} \right\}$$
$$ = {\,^{21}}{C_4} - {\,^{10}}{C_4}$$

$$\eqalign{ & \frac{{{2^{4n}}}}{{15}} = \frac{{{{16}^n}}}{{15}} = \frac{{{{\left( {1 + 15} \right)}^n}}}{{15}} \cr & = \frac{{1 + {\,^n}{C_1} 15 + {\,^n}{C_2} {{15}^2} + ..... + {\,^n}{C_n} {{15}^n}}}{{15}} \cr & = \frac{{1 + 15k}}{{15}},{\text{where }}k \in N, = \frac{1}{{15}} + k \cr} $$
$$\therefore $$ Fractional part of $$\frac{{{2^{4n}}}}{{15}}\,\,{\text{is}}\,\,\frac{1}{{15}}$$

Take $$n = 2m.$$  Then
$$\eqalign{ & {a_n} = \frac{1}{{^{2m}{C_0}}} + \frac{1}{{^{2m}{C_1}}} + ..... + \frac{1}{{^{2m}{C_{2m}}}} \cr & {a_n} = 2\left( {\frac{1}{{^{2m}{C_0}}} + \frac{1}{{^{2m}{C_1}}} + ..... + \frac{1}{{^{2m}{C_{m - 1}}}}} \right) + \frac{1}{{^{2m}{C_m}}}\,\,\,\,.....\left( 1 \right) \cr & \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} = \sum\limits_{r = 0}^{2m} {\frac{r}{{^{2m}{C_r}}} = \frac{1}{{^{2m}{C_1}}} + \frac{2}{{^{2m}{C_2}}} + ..... + \frac{{2m}}{{^{2m}{C_{2m}}}}} \cr & \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} = \left( {\frac{1}{{^{2m}{C_1}}} + \frac{{2m - 1}}{{^{2m}{C_{2m - 1}}}}} \right) + \left( {\frac{2}{{^{2m}{C_2}}} + \frac{{2m - 2}}{{^{2m}{C_{2m - 2}}}}} \right) + ..... + \left( {\frac{{m - 1}}{{^{2m}{C_{m - 1}}}} + \frac{{m + 1}}{{^{2m}{C_{m + 1}}}}} \right) + \frac{m}{{^{2m}{C_m}}} + \frac{{2m}}{{^{2m}{C_{2m}}}} \cr & \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} = 2m\left\{ {\frac{1}{{^{2m}{C_1}}} + \frac{1}{{^{2m}{C_2}}} + ..... + \frac{1}{{^{2m}{C_{m - 1}}}}} \right\} + \frac{m}{{^{2m}{C_m}}} + 2m \cr & \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} = 2m\left\{ {\frac{1}{{^{2m}{C_0}}} + \frac{1}{{^{2m}{C_1}}} + \frac{1}{{^{2m}{C_2}}} + ..... + \frac{1}{{^{2m}{C_{m - 1}}}}} \right\} + \frac{m}{{^{2m}{C_m}}} \cr & \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} = m\left( {{a_n} - \frac{1}{{^{2m}{C_m}}}} \right) + \frac{m}{{^{2m}{C_m}}} = m{a_n} = \frac{n}{2}{a_n}. \cr} $$
Similarly for $$n = 2m + 1.$$

Putting $$x = i, - i$$   and multiplying both the results, we get the value of the required expression.

$$\eqalign{ & {\left( {x + a} \right)^n} = {\,^n}{C_0}{x^n} + {\,^n}{C_1}{x^{n - 1}}a + {\,^n}{C_2}{x^{n - 2}}{a^2} + {\,^n}{C_3}{x^{n - 3}}{a^3} + {\,^n}{C_4}{x^{n - 4}}{a^4} + ..... \cr & = \left( {^n{C_0}{x^n} + {\,^n}{C_2}{x^{n - 2}}{a^2} + {\,^n}{C_4}{x^{n - 4}}{a^4} + .....} \right) + \left( {^n{C_1}{x^{n - 1}}a + {\,^n}{C_3}{x^{n - 3}}{a^3} + {\,^n}{C_5}{x^{n - 5}}{a^5}} \right) + ..... \cr & = A + B\,\,\,\,.....\left( 1 \right) \cr} $$
Similarly, $${\left( {x - a} \right)^n} = A - B\,\,\,\,.....\left( 2 \right)$$
Multiplying eqns. (1) and (2), we get
$${\left( {{x^2} - {a^2}} \right)^n} = {A^2} - {B^2}$$