Binomial Theorem MCQ Questions & Answers in Algebra | Maths

\[\begin{array}{l} \left( {\begin{array}{*{20}{c}} n\\ r \end{array}} \right) + 2\left( {\begin{array}{*{20}{c}} n\\ {r - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {r - 2} \end{array}} \right)\\ = \left[ {\left( {\begin{array}{*{20}{c}} n\\ r \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {r - 1} \end{array}} \right)} \right] + \left[ {\left( {\begin{array}{*{20}{c}} n\\ {r - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {r - 2} \end{array}} \right)} \right] \end{array}\]
NOTE THIS STEP :
\[\left( {\begin{array}{*{20}{c}} {n + 1}\\ r \end{array}} \right) + \left( {\begin{array}{*{20}{c}} {n + 1}\\ {r - 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {n + 2}\\ r \end{array}} \right)\]
$$\left[ {\because \,{\,^n}{C_r} + {\,^n}{C_{r - 1}} = {\,^{n + 1}}{C_r}} \right]$$

$$\eqalign{ & ^{n - 1}{C_r} = {\,^n}{C_{r + 1}}\left( {{k^2} - 3} \right) \cr & \Rightarrow \,\,{k^2} - 3 = \frac{{^{n - 1}{C_r}}}{{^n{C_{r + 1}}}} = \frac{{r + 1}}{n} \cr & {\text{Since }}0 \leqslant r \leqslant n - 1 \cr & \Rightarrow \,\,1 \leqslant r + 1 \leqslant n \cr & \Rightarrow \,\,\frac{1}{n} \leqslant \frac{{r + 1}}{n} \leqslant 1 \cr & \Rightarrow \,\,\frac{1}{n} \leqslant {k^2} - 3 \leqslant 1 \cr & \Rightarrow \,\,3 + \frac{1}{n} \leqslant {k^2} \leqslant 4 \cr & \Rightarrow \,\,\sqrt {3 + \frac{1}{n}} \leqslant k \leqslant 2 \cr & {\text{as }}\,n \to \infty \cr & \Rightarrow \,\,\sqrt 3 < k \leqslant 2 \cr & \Rightarrow \,\,k \in \left( {\sqrt 3 ,2} \right] \cr} $$

To find
$$^{30}{C_0}^{30}{C_{10}} - {\,^{30}}{C_1}^{30}{C_{11}} + {\,^{30}}{C_2}^{30}{C_{12}} - .....{ + ^{30}}{C_{20}}^{30}{C_{30}}$$
We know that
$$\eqalign{ & {\left( {1 + x} \right)^{30}}{ = ^{30}}{C_0} + {\,^{30}}{C_1}x + {\,^{30}}{C_2}{x^2} + ..... + {\,^{30}}{C_{20}}{x^{20}} + .....{\,^{30}}{C_{30}}{x^{30}}\,\,\,\,\,\,.....\left( 1 \right) \cr & {\left( {x - 1} \right)^{30}} = {\,^{30}}{C_0}{x^{30}} - {\,^{30}}{C_1}{x^{29}} + ..... + {\,^{30}}{C_{10}}{x^{20}} - {\,^{30}}{C_{11}}{x^{19}} + {\,^{30}}{C_{12}}{x^{18}} + {.....^{30}}{C_{30}}{x^0}\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
Multiplying $${\text{e}}{{\text{q}}^n}$$ (1) and (2), we get
$${\left( {{x^2} - 1} \right)^{30}} = \left( {} \right) \times \left( {} \right)$$
Equating the co - efficients of $${x^{20}}$$ on both sides, we get
$$^{30}{C_{10}} = {\,^{30}}{C_0}^{30}{C_{10}} - {\,^{30}}{C_1}^{30}{C_{11}} + {\,^{30}}{C_2}^{30}{C_{12}} - ..... + {\,^{30}}{C_{20}}^{30}{C_{30}}$$
∴ Req. value is $$^{30}{C_{10}}$$

$$\eqalign{ & x = {\left( {2 + \sqrt 3 } \right)^n} \cr & = {\,^n}{C_0}{2^n} + {\,^n}{C_1}{2^{n - 1}}\sqrt 3 + {\,^n}{C_2}{2^{n - 2}}{\left( {\sqrt 3 } \right)^2} + ..... \cr & {\text{Let, }}{x_1} = {\left( {2 - \sqrt 3 } \right)^n} \cr & = {\,^n}{C_0}{2^n} - {\,^n}{C_1}{2^{n - 1}}\sqrt 3 + {\,^n}{C_2}{2^{n - 2}}{\left( {\sqrt 3 } \right)^2} + ..... \cr & x + {x_1} = 2\left( {^n{C_0}{2^n} + {\,^n}{C_2}{2^{n - 2}}{{\left( {\sqrt 3 } \right)}^2} + .....} \right) \cr} $$
= Even integer.
Clearly, $$x, \in \left( {0,1} \right)\forall n \in N$$
$$\eqalign{ & \Rightarrow \left[ x \right] + \left\{ x \right\} + {x_1} = {\text{Even integer}} \cr & \Rightarrow \left\{ x \right\} + {x_1} = {\text{Integer }}\left\{ x \right\} \in \left( {0,1} \right),{x_1} \in \left( {0,1} \right) \cr & \Rightarrow \left\{ x \right\} + {x_1} \in \left( {0,2} \right) \cr & \Rightarrow \left\{ x \right\} + {x_1} = 1 \cr & \Rightarrow {x_1} = 1 - \left\{ x \right\} \cr & \Rightarrow x\left( {1 - \left\{ x \right\}} \right) = x \cdot {x_1} = {\left( {2 + \sqrt 3 } \right)^n}{\left( {2 - \sqrt 3 } \right)^n} = 1 \cr} $$

$${\left( {x + y - z} \right)^{16}} = {\,^{16}}{C_0}{x^{16}} + {\,^{16}}{C_1}{x^{15}}\left( {y - z} \right) + ..... + {\,^{16}}{C_r}{x^{16 - r}}{\left( {y - z} \right)^r} + ..... + {\,^{16}}{C_{16}}{\left( {y - z} \right)^{16}}.$$
Clearly, all the terms are distinct.
∴ the number of distinct terms
$$ = 1 + 2 + 3 + ..... + 17 = \frac{{17 \times 18}}{2}.$$

$$\eqalign{ & \frac{{\pi \left( n \right)}}{{\pi \left( {n + 1} \right)}} = \frac{{^n{C_0} \cdot {\,^n}{C_1} \cdot {\,^n}{C_2}{{.....}^n}{C_n}}}{{^{n + 1}{C_0} \cdot {\,^{n + 1}}{C_1} \cdot {\,^{n + 1}}{C_2}{{.....}^{n + 1}}{C_{n + 1}}}} \cr & = \frac{1}{{^{n + 1}{C_0}}}\left( {\frac{{^n{C_0}}}{{^{n + 1}{C_1}}}} \right)\left( {\frac{{^n{C_1}}}{{^{n + 1}{C_2}}}} \right).....\left( {\frac{{^n{C_n}}}{{^{n + 1}{C_{n + 1}}}}} \right) \cr & = \frac{1}{1}\left( {\frac{1}{{n + 1}}} \right)\left( {\frac{2}{{n + 1}}} \right).....\left( {\frac{{n + 1}}{{n + 1}}} \right)\left[ {\because \frac{{^n{C_r}}}{{^{n + 1}{C_{r + 1}}}} = \frac{{r + 1}}{{n + 1}}} \right] \cr & = \frac{{\left( {n + 1} \right)!}}{{{{\left( {n + 1} \right)}^{n + 1}}}} = \frac{{n!}}{{{{\left( {n + 1} \right)}^n}}} \cr & \therefore \frac{{\pi \left( {2002} \right)}}{{\pi \left( {2001} \right)}} = \frac{{{{\left( {2002} \right)}^{2001}}}}{{\left( {2001} \right)!}} \cr} $$

The sum of all the co-efficients in an expansion is obtained by putting $$x = 1$$  in the expression.
$$\eqalign{ & \therefore \,\,{\left( {\frac{1}{1} + 2 \cdot 1} \right)^n} = 6561 \cr & \therefore \,\,{3^n} = {3^8}\,\,\,\,\,\,\,\therefore \,\,n = 8. \cr & {\text{In}}{\left( {\frac{1}{x} + 2x} \right)^8},{t_{r + 1}} = {\,^8}{C_r} \cdot {\left( {\frac{1}{x}} \right)^{8 - r}} \cdot {\left( {2x} \right)^r}. \cr} $$
This is a constant if $$2r - 8 = 0,\,\,{\text{i}}{\text{.e}}{\text{., }}r = 4.$$
∴ the constant term $$ = {t_5} = {\,^8}{C_4} \cdot {2^4}.$$

$$\eqalign{ & \frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + ..... = \frac{n}{2} + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{4!}} + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)}}{{6!}} + ..... \cr & = \frac{1}{{n + 1}}\left[ {\frac{{\left( {n + 1} \right)n}}{{2!}} + \frac{{\left( {n + 1} \right)n\left( {n - 1} \right)\left( {n - 2} \right)}}{{4!}} + \frac{{\left( {n + 1} \right)n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)}}{{6!}} + .....} \right] \cr & = \frac{1}{{n + 1}}\left[ {^{n + 1}{C_2} + {\,^{n + 1}}{C_4} + {\,^{n + 1}}{C_6} + .....} \right] \cr & = \frac{1}{{n + 1}}\left[ {{2^{n + 1 - 1}} - {\,^{n + 1}}{C_0}} \right] = \frac{{{2^n} - 1}}{{n + 1}} \cr} $$

$${2^{60}} = {\left( {1 + 7} \right)^{20}} = {\,^{20}}{C_0} + {\,^{20}}{C_1} \cdot 7 + {\,^{20}}{C_2} \cdot {7^2} + ..... + {\,^{20}}{C_{20}} \cdot {7^{20}}$$
∴ the remainder $$ = {\,^{20}}{C_0} = 1.$$

$$\because {x^3}$$  and higher powers of $$x$$ may be neglected
$$\eqalign{ & \therefore \frac{{{{\left( {1 + x} \right)}^{\frac{3}{2}}} - {{\left( {1 + \frac{x}{2}} \right)}^3}}}{{\left( {1 - {x^{\frac{1}{2}}}} \right)}} \cr & = {\left( {1 - x} \right)^{ - \frac{1}{2}}}\left[ {\left( {1 + \frac{3}{2}x + \frac{{\frac{3}{2} \cdot \frac{1}{2}}}{{2!}}{x^2}} \right) - \left( {1 + \frac{{3x}}{2} + \frac{{3 \cdot 2}}{{2!}}\frac{{{x^2}}}{4}} \right)} \right] \cr & = \left[ {1 + \frac{x}{2} + \frac{{\frac{1}{2} \cdot \frac{3}{2}}}{{2!}}{x^2}} \right]\left[ { - \frac{3}{8}{x^2}} \right] = - \frac{3}{8}{x^2} \cr} $$
(as $$x^3$$ and higher powers of $$x$$ can be neglected)