Binomial Theorem MCQ Questions & Answers in Algebra | Maths

The general term of the series
$$\eqalign{ & \frac{x}{{2!}} + \frac{{2{x^2}}}{{3!}} + \frac{{3{x^3}}}{{4!}} + .....\,\infty {\text{ is}} \cr & {T_n} = \frac{{n{x^n}}}{{\left( {n + 1} \right)!}},n = 1,2,.....,\infty \cr & = \frac{{n + 1 - 1}}{{\left( {n + 1} \right)!}}{x^n} = \frac{{{x^n}}}{{n!}} - \frac{1}{x}\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}} \cr & \therefore 1 + \frac{x}{{2!}} + \frac{{2{x^2}}}{{3!}} + \frac{{3{x^3}}}{{4!}} + .....\,\infty \cr & = 1 + \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n!}} - \frac{1}{x}\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}}} } \cr & = 1 + \left( {{e^x} - 1} \right) - \frac{1}{x}\left( {{e^x} - 1 - x} \right) \cr & = \frac{{x{e^x} - {e^x} + 1 + x}}{x} = \frac{{\left( {x - 1} \right){e^x} + \left( {1 + x} \right)}}{x} \cr} $$

$$\eqalign{ & 1 + {99^n} = 1 + {\left( {100 - 1} \right)^n} = 1 + \left\{ {^n{C_0}{{100}^n} - {\,^n}{C_1} \cdot {{100}^{n - 1}} + ..... - {\,^n}{C_n}} \right\}\,\,{\text{because }}n{\text{ is odd}} \cr & 1 + {99^n} = 100\left\{ {^n{C_0} \cdot {{100}^{n - 1}} - {\,^n}{C_1} \cdot {{100}^{n - 2}} + ..... - {\,^n}{C_{n - 2}} \cdot 100 + {\,^n}{C_{n - 1}}} \right\} \cr} $$
$$1 + {99^n} = 100 \times $$    integer whose units place is different from 0 ( $$n$$ having odd digit in units place).

$$\eqalign{ & {\left( {\frac{a}{{a + x}}} \right)^{\frac{1}{2}}} + {\left( {\frac{a}{{a - x}}} \right)^{\frac{1}{2}}} \cr & = {\left( {\frac{{a + x}}{a}} \right)^{ - \frac{1}{2}}} + {\left( {\frac{{a - x}}{a}} \right)^{ - \frac{1}{2}}} \cr & = {\left( {1 + \frac{x}{a}} \right)^{ - \frac{1}{2}}} + {\left( {1 - \frac{x}{a}} \right)^{ - \frac{1}{2}}} \cr & = \left[ {1 - \frac{1}{2}\frac{x}{a} + \frac{3}{8}\frac{{{x^2}}}{{{a^2}}}} \right] + \left[ {1 + \frac{1}{2}\frac{x}{a} + \frac{3}{8}\frac{{{x^2}}}{{{a^2}}}} \right]\left[ {\because x \ll a,\therefore \frac{x}{a} \ll 1} \right] = 2 + \frac{3}{4} \cdot \frac{{{x^2}}}{{{a^2}}} \cr} $$

$${t_{r + 1}} = {\,^{600}}{C_r} \cdot {7^{\frac{{600 - r}}{3}}} \cdot {5^{\frac{r}{2}}}{x^r}.$$
Here, $$0 \leqslant r \leqslant 600$$   and $$\frac{r}{2},200 - \frac{r}{3}$$   are integers.
∴ $$r$$ should be a multiple of 6
$$\therefore \,\,r = 0,6,12,.....,600.$$

Sum $$ = a\sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^{r - 1}} \cdot {\,^n}{C_r}} - \sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^{r - 1}} \cdot r \cdot {\,^n}{C_r}} $$
$$\eqalign{ &\,\,\,\,\,\,\,\,\,\,\,\,\, = a\left\{ {^n{C_1} - {\,^n}{C_2} + {\,^n}{C_3} - .....} \right\} - n\left\{ {^{n - 1}{C_0} - {\,^{n - 1}}{C_1} + .....} \right\}\left( {{\text{because }}r \cdot {\,^n}{C_r} = n \cdot {\,^{n - 1}}{C_{r - 1}}} \right) \cr &\,\,\,\,\,\,\,\,\,\,\,\,\, = a \cdot {\,^n}{C_0}\,\,\left( {\because \,{\,^n}{C_0} - {\,^n}{C_1} + {\,^n}{C_2} - ..... = 0} \right). \cr} $$

Expression $$ = \frac{{^n{C_0} + {\,^n}{C_1}}}{{{\,^n}{C_0}}} \cdot \frac{{{\,^n}{C_1} + {\,^n}{C_2}}}{{{\,^n}{C_1}}}.....\frac{{{\,^n}{C_{n - 1}} + {\,^n}{C_n}}}{{{\,^n}{C_{n - 1}}}}$$
$$\eqalign{ & = \frac{{{\,^{n + 1}}{C_1}}}{{{\,^n}{C_0}}} \cdot \frac{{{\,^{n + 1}}{C_2}}}{{{\,^n}{C_1}}}.....\frac{{{\,^{n + 1}}{C_n}}}{{{\,^n}{C_{n - 1}}}} \cr & = \frac{{1 \cdot {\,^{n + 1}}{C_1}}}{{{\,^n}{C_0}}} \cdot \frac{{2 \cdot {\,^{n + 1}}{C_2}}}{{{\,^n}{C_1}}}.....\frac{{n \cdot {\,^{n + 1}}{C_n}}}{{{\,^n}{C_{n - 1}}}} \cdot \frac{1}{{n!}} \cr & = \frac{{\left( {n + 1} \right) \cdot {\,^n}{C_0}}}{{{\,^n}{C_0}}} \cdot \frac{{\left( {n + 1} \right) \cdot {\,^n}{C_1}}}{{{\,^n}{C_1}}}.....\frac{{\left( {n + 1} \right) \cdot {\,^n}{C_{n - 1}}}}{{{\,^n}{C_{n - 1}}}} \cdot \frac{1}{{n!}}. \cr} $$

$${T_{r + 1}} = \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)......\left( {n - r + 1} \right)}}{{r!}}{\left( x \right)^r}$$
For first negative term, $$n - r + 1 < 0$$
$$ \Rightarrow \,\,r > n + 1$$
$$ \Rightarrow \,\,r > \frac{{32}}{5}\,\,\,\,\,\therefore \,\,r = 7.\left( {\because \,\,n = \frac{{27}}{5}} \right)$$
Therefore, first negative term is $${T_8}.$$

The co-efficients of the integral powers of $$x$$ are $$^{40}{C_0},{\,^{40}}{C_2} \cdot {2^2},{\,^{40}}{C_4} \cdot {2^4},.....,{\,^{40}}{C_{40}} \cdot {2^{40}}.$$
$$\eqalign{ & {\left( {1 + 2} \right)^{40}} = {\,^{40}}{C_0} + {\,^{40}}{C_1} \cdot 2 + {\,^{40}}{C_2} \cdot {2^2} + ..... + {\,^{40}}{C_{40}} \cdot {2^{40}}. \cr & {\left( {1 - 2} \right)^{40}} = {\,^{40}}{C_0} - {\,^{40}}{C_1} \cdot 2 - {\,^{40}}{C_2} \cdot {2^2} - ..... + {\,^{40}}{C_{40}} \cdot {2^{40}}. \cr} $$
Adding, $${3^{40}} + 1 = 2 \times \left( {{\text{required sum}}} \right).$$

$$\eqalign{ & \sqrt 5 \left[ {{{\left( {\sqrt 5 + 1} \right)}^{50}} - {{\left( {\sqrt 5 - 1} \right)}^{50}}} \right] \cr & = 2\sqrt 5 \left[ {^{50}{C_1}{{\left( {\sqrt 5 } \right)}^{49}} + {\,^{50}}{C_3}{{\left( {\sqrt 5 } \right)}^{47}} + .....} \right] \cr & = 2\left[ {^{50}{C_1}{{\left( {\sqrt 5 } \right)}^{50}} + {\,^{50}}{C_3}{{\left( {\sqrt 5 } \right)}^{48}} + .....} \right] \cr} $$
= a natural number

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