Binomial Theorem MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & {\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n} \cr & = \left[ {1 - {\,^m}{C_1}y + {\,^m}{C_2}{y^2} - .....} \right]\left[ {1 + {\,^n}{C_1}y + {\,^n}{C_2}{y^2} + .....} \right] \cr & = 1 + \left( {n - m} \right) + \left\{ {\frac{{m\left( {m - 1} \right)}}{2} + \frac{{n\left( {n - 1} \right)}}{2} - mn} \right\}{y^2} + ..... \cr & \therefore \,\,{a_1} = n - m = 10 \cr & {\text{and }}{a_2} = \frac{{{m^2} + {n^2} - m - n - 2mn}}{2} = 10 \cr & {\text{So, }}n - m = 10\,\,{\text{and }}{\left( {m - n} \right)^2} - \left( {m + n} \right) = 20 \cr & \Rightarrow \,\,m + n = 80 \cr & \therefore \,\,m = 35,\,\,n = 45. \cr} $$

We have, $$y = 3x + 6{x^2} + 10{x^3} + .....$$
$$\eqalign{ & \Rightarrow 1 + y = 1 + 3x + 6{x^2} + 10{x^3} + ..... \cr & \Rightarrow 1 + y = {\left( {1 - x} \right)^{ - 3}} \cr & \Rightarrow 1 - x = {\left( {1 + y} \right)^{ - \frac{1}{3}}} \cr & \Rightarrow x = 1 - {\left( {1 + y} \right)^{ - \frac{1}{3}}} \cr & = \frac{1}{3}y - \frac{{1 \cdot 4}}{{{3^2} \cdot 2}}{y^2} + \frac{{1 \cdot 4 \cdot 7}}{{{3^2} \cdot 3}}{y^3} - ..... \cr} $$

Suppose $$x^{- 7}$$ occurs in $${\left( {r + 1} \right)^{th}}{\text{term}}{\text{.}}$$
we have $${T_{r + 1}} = {\,^n}{C_r}{x^{n - r}}{a^r}{\text{ in }}{\left( {x + a} \right)^n}.$$
In the given question, $$n = 11,x = ax,a = \frac{{ - 1}}{{b{x^2}}}$$
$$\eqalign{ & \therefore {T_{r + 1}} = {\,^{11}}{C_r}{\left( {ax} \right)^{11 - r}}{\left( {\frac{{ - 1}}{{b{x^2}}}} \right)^r} \cr & = {\,^{11}}{C_r}{a^{11 - r}}{b^{ - r}}{x^{11 - 3r}}{\left( { - 1} \right)^r} \cr} $$
This term contains $$x^{- 7}$$ if $$11 - 3r = - 7$$
$$ \Rightarrow r = 6$$
Therefore, coefficient of $$x^{ - 7}$$ is
$$^{11}{C_6}{\left( a \right)^5}{\left( {\frac{{ - 1}}{b}} \right)^6} = \frac{{462}}{{{b^6}}}{a^5}$$

Sum $$ = \frac{1}{2}\left\{ {\left( {{a_0} + {a_1} + {a_2} + ..... + {a_{16}}} \right) - \left( {{a_0} - {a_1} + {a_2} - ..... + {a_{16}}} \right)} \right\}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left\{ {{{\left( {1 + 1 - 2} \right)}^8} - {{\left( {1 - 1 - 2} \right)}^8}} \right\} = \frac{1}{2}\left( { - {2^8}} \right) = - {2^7}.$$

$$\eqalign{ & \sum\limits_{r = 0}^n {\frac{{r + 2}}{{r + 1}}{\,^n}{C_r} = \sum\limits_{r = 0}^n {\frac{{r + 1 + 1}}{{r + 1}}{\,^n}{C_r}} } \cr & = \sum\limits_{r = 0}^n {^n{C_r}} + \sum\limits_{r = 0}^n {\frac{{^n{C_r}}}{{r + 1}} = {2^n} + \sum\limits_{r = 0}^n {\frac{{^{n + 1}{C_{r + 1}}}}{{n + 1}}} } \cr & = {2^n} + \frac{1}{{n + 1}}\sum\limits_{r = 0}^n {^{n + 1}{C_{r + 1}}} \cr & = {2^n} + \frac{1}{{n + 1}}\left( {{2^{n + 1}} - 1} \right) \cr & = \frac{1}{{n + 1}}\left[ {\left( {n + 1} \right){2^n} + {2^{n + 1}} - 1} \right] \cr & = \frac{1}{{n + 1}}\left[ {{2^n}\left( {n + 3} \right) - 1} \right] \cr & {\text{Given, }}\frac{{\left( {n + 3} \right){2^n} - 1}}{{n + 1}} = \frac{{{2^8} - 1}}{6} = \frac{{\left( {5 + 3} \right) \cdot {2^5} - 1}}{{5 + 1}} \cr & \Rightarrow n = 5 \cr} $$

Let the given series be the expansion of $${\left( {1 + x} \right)^n},$$   then it is identical with
$$\eqalign{ & 1 + nx + \frac{{n\left( {n - 1} \right)}}{{2!}} \cdot {x^2} + ..... \cr & \therefore \,nx = \frac{1}{3}\,\,\,.....\left( 1 \right) \cr & \frac{{n\left( {n - 1} \right)}}{2} \cdot {x^2} = \frac{1}{6}\,\,\,.....\left( 2 \right) \cr} $$
Solving the equations (1) and (2)
we get $$n = - \frac{1}{2}{\text{ and }}x = - \frac{2}{3}$$
$$\therefore $$ The given series
$$ = \,{\left( {1 - \frac{2}{3}} \right)^{ - \frac{1}{2}}} = {\left( {\frac{1}{3}} \right)^{ - \frac{1}{2}}} = \sqrt 3 $$

Using $$r \cdot {\,^n}{C_r} = n \cdot {\,^{n - 1}}{C_{r - 1}},$$
sum $$ = 20\left\{ {^{19}{C_0} - {\,^{19}}{C_1} + {\,^{19}}{C_2} - ..... - {\,^{19}}{C_{19}}} \right\} = 20 \times 0 = 0.$$

$$\eqalign{ & \frac{{\left( {3 - 2x} \right)}}{{{{\left( {1 + 3x} \right)}^3}}} = \left( {3 - 2x} \right){\left( {1 + 3x} \right)^{ - 3}} \cr & = \left( {3 - 2x} \right)\left[ {1 - 9x + \frac{{\left( { - 3} \right)\left( { - 4} \right)}}{{2!}} \cdot 9{x^2} + \frac{{\left( { - 3} \right)\left( { - 4} \right)\left( { - 5} \right)}}{{3!}} \cdot 27{x^3} + .....} \right] \cr & \left[ {{\text{Expanding}}\,{{\left( {1 + 3x} \right)}^{ - 3}}} \right] \cr & = \left( {3 - 2x} \right)\left( {1 - 9x + 54{x^2} - 270{x^3} + .....} \right) \cr & \therefore {\text{Coefficient of }}{x^3} = - 270 \times 3 - 2 \times 54 = - 918 \cr} $$

$$\eqalign{ & {\left( {1 + 0.0001} \right)^{10000}} = {\left( {1 + \frac{1}{n}} \right)^n},n = 10000 \cr & = 1 + n \cdot \frac{1}{n} + \frac{{n\left( {n - 1} \right)}}{{2!}}\frac{1}{{{n^2}}} + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}\frac{1}{{{n^3}}} + ..... \cr & = 1 + 1 + \frac{1}{{2!}}\left( {1 - \frac{1}{n}} \right) + \frac{1}{{3!}}\left( {1 - \frac{1}{n}} \right)\left( {1 - \frac{2}{n}} \right) + ..... < 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + ..... + \frac{1}{{\left( {9999} \right)!}} \cr & = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + .....\,\infty = e < 3 \cr} $$

Total number of terms $$ = {\,^{n + 2}}{C_2} = 28$$
$$\eqalign{ & \left( {n + 2} \right)\left( {n + 1} \right) = 56 \cr & x = 6 \cr} $$
Sum of co-efficients
$$\eqalign{ & = {\left( {1 - 2 + 4} \right)^n} \cr & = {3^6} \cr & = 729 \cr} $$