Binomial Theorem MCQ Questions & Answers in Algebra | Maths

Put $$x = y = z = 1,$$    the sum of coefficient $$ = {\left( {1 + 2 + 3} \right)^{10}} = {6^{10}}.$$

Expression $$ = {\left( {1 - x} \right)^5} \cdot {\left( {1 + x} \right)^4}{\left( {1 + {x^2}} \right)^4} = \left( {1 - x} \right){\left( {1 - {x^2}} \right)^4}{\left( {1 + {x^2}} \right)^4}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {1 - x} \right){\left( {1 - {x^4}} \right)^4}.$$
∴ the co-efficient of $${x^{13}} = - {\,^4}{C_3}{\left( { - 1} \right)^3} = 4.$$

We have,
$$\eqalign{ & {\left\{ {{3^{{{\log }_3}\sqrt {{{25}^{x - 1}} + 7} }} + {3^{ - \frac{1}{8}{{\log }_3}\left( {{5^{x - 1}} + 1} \right)}}} \right\}^{10}} \cr & = {\left[ {\sqrt {{{25}^{x - 1}} + 7} + {{\left( {{5^{x - 1}} + 1} \right)}^{ - \frac{1}{8}}}} \right]^{10}}\left( {{\text{since, }}{a^{{{\log }_a}N}} = N} \right) \cr & {\text{Here, }}{T_9} = 180 \cr & \Rightarrow {\,^{10}}{C_8}{\left\{ {\sqrt {{{25}^{x - 1}} + 7} } \right\}^{10 - 8}}{\left\{ {{{\left( {{5^{x - 1}} + 1} \right)}^{ - \frac{1}{8}}}} \right\}^8} = 180 \cr & \Rightarrow {\,^{10}}{C_8}\left( {{{25}^{x - 1}} + 7} \right){\left( {{5^{x - 1}} + 1} \right)^{ - 1}} = 180 \cr & \Rightarrow 45\frac{{\left( {{{25}^{x - 1}} + 7} \right)}}{{{5^{x - 1}} + 1}} = 180 \cr & \Rightarrow \frac{{{{25}^{x - 1}} + 7}}{{{5^{x - 1}} + 1}} = 4 \cr & \Rightarrow \frac{{{y^2} + 7}}{{y + 1}} = 4,{\text{where }}y = {5^{x - 1}} \cr & \Rightarrow {y^2} - 4y + 3 = 0 \cr & \Rightarrow y = 3,1 \cr & \Rightarrow {5^{x - 1}} = 3{\text{ or }}{5^{x - 1}} = 1 \cr & \Rightarrow {5^x} = 15{\text{ or }}{5^x} = 5 \cr & \Rightarrow x = {\log _5}15{\text{ or }}x = 1 \cr} $$

$${t_{r + 1}} = {\,^{10}}{C_r}{\left( {x\sin \alpha } \right)^{10 - r}} \cdot {\left( {\frac{{\cos \alpha }}{x}} \right)^r}.$$
It is independent of $$x$$ if $$r = 5.$$
∴ the term independent of $$x = {\,^{10}}{C_5} \cdot {\sin ^5}\alpha \cdot {\cos ^5}\alpha $$
$$x = {\,^{10}}{C_5} \cdot \frac{1}{{{2^5}}}{\left( {\sin 2\alpha } \right)^5} \leqslant {\,^{10}}{C_5} \cdot \frac{1}{{{2^5}}}.$$

Middle term in the expansion is $${\left( {\frac{{10}}{2} + 1} \right)^{th}}{\text{ i}}{\text{.e}}{\text{.,}}\,{6^{th}}{\text{term}}{\text{.}}$$
Thus, $${T_6} = 7\frac{7}{8}$$
$$\eqalign{ & \Rightarrow {\,^{10}}{C_5}\frac{1}{{{x^5}}} \cdot {x^5}{\sin ^5}x = \frac{{63}}{8} \cr & \Rightarrow 252 \cdot {\sin ^5}x = \frac{{63}}{8} \cr & \Rightarrow {\sin ^5}x = \frac{1}{{32}} \cr & \Rightarrow \sin x = \frac{1}{2} \cr & \therefore x = n\pi + {\left( { - 1} \right)^n}\frac{\pi }{6} \cr} $$

Sum $$ = \frac{1}{2}\,\left\{ {^{10}{C_0} - {\,^{10}}{C_1} \cdot 2 + {\,^{10}}{C_2} \cdot {2^2} - {\,^{10}}{C_3} \cdot {2^3} + ..... + {\,^{10}}{C_{10}} \cdot {2^{10}}} \right\}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}{\left( {1 - 2} \right)^{10}} = \frac{1}{2}.$$

If $$n$$ is odd, then numerically greatest coefficient in the expansion of $${\left( {1 - x} \right)^n}\,$$  is $$\frac{{^n{C_{n - 1}}}}{2}\,{\text{or}}\,\frac{{^n{C_{n + 1}}}}{2}.$$
Therefore in $${\left( {1 - x} \right)^{21}},\,$$  the numerically greatest coefficient is $$^{21}{C_{10}}$$  or $$^{21}{C_{11}}.$$  So, the numerically greatest term
$$\eqalign{ & = {\,^{21}}{C_{11}}{x^{11}}{\text{ or}}{\,^{21}}{C_{10}}{x^{10}} \cr & {\text{So, }}\left| {^{21}{C_{11}}{x^{11}}} \right| > \left| {^{21}{C_{12}}{x^{12}}} \right|{\text{ and}} \cr & \left| {^{21}{C_{10}}{x^{10}}} \right| > \left| {^{21}{C_9}{x^9}} \right| \cr & \Rightarrow \frac{{21!}}{{10!\,11!}} > \frac{{21!}}{{9!\,12!}}x{\text{ and }}\frac{{21!}}{{11!\,10!}}x > \frac{{21!}}{{9!\,12!}} \cr & \left( {\because x > 0} \right) \cr & \Rightarrow x < \frac{6}{5}{\text{ and }}x > \frac{5}{6} \cr & \Rightarrow x \in \left( {\frac{5}{6},\frac{6}{5}} \right) \cr} $$

$$\eqalign{ & {\left( {1.0002} \right)^{3000}} = {\left( {1 + 0.0002} \right)^{3000}} \cr & = 1 + \left( {3000} \right)\left( {0.0002} \right) + \frac{{\left( {3000} \right)\left( {2999} \right)}}{{1.2}}{\left( {0.0002} \right)^2} + \frac{{\left( {3000} \right)\left( {2999} \right)\left( {2998} \right)}}{{1.2.3}}{\left( {0.0002} \right)^3} + ..... \cr} $$
We want to get answer correct to only one decimal places and as such we have left further expansion.
$$ = 1 + \left( {3000} \right)\left( {0.0002} \right) = 1.6$$

Given expansion is $${\left( {a + bx} \right)^{ - 3}}$$   which can be written as
$$\eqalign{ & {\left[ {a\left( {1 + \frac{b}{a}x} \right)} \right]^{ - 3}} = {a^{ - 3}}{\left( {1 + \frac{b}{a}x} \right)^{ - 3}} \cr & = {a^{ - 3}}\left( {1 - \frac{{3b}}{a}x + 6{{\left( {\frac{b}{a}x} \right)}^2} - .....} \right) \cr & \left( {{\text{By using}}{{\left( {1 + x} \right)}^{ - 3}} = 1 - 3x + 6{x^2} - .....} \right) \cr} $$
But given that : $${\left( {a + bx} \right)^{ - 3}} = \frac{1}{8} + \frac{9}{8}x + .....$$
$$\eqalign{ & \therefore {a^{ - 3}}\left[ {1 - \frac{{3b}}{a}x + 6\frac{{{b^2}}}{{{a^2}}} \cdot {x^2} - .....} \right] = \frac{1}{8} + \frac{9}{8}x + ..... \cr & \Rightarrow {a^{ - 3}} = \frac{1}{8} = {2^{ - 3}} \cr & \Rightarrow a = 2 \cr & {\text{and }} - 3b{a^{ - 4}} = 9 \cdot {2^{ - 3}} \cr & \Rightarrow b = - 6 \cr} $$

The greatest co-efficient
= the coefficient of the middle term
$$ = {\,^{2n}}{C_n}.$$
Now simplify.