Binomial Theorem MCQ Questions & Answers in Algebra | Maths

$${\left( {x\sin p + {x^{ - 1}}\cos p} \right)^{10}},$$     general term is $${T_{r + 1}} = {\,^{10}}{C_r}{\left( {x\sin p} \right)^{10 - r}}{\left( {{x^{ - 1}}\cos p} \right)^r}.$$
For the term independent of $$x$$ we have $$10 - 2r = 0$$   or $$r = 5$$
Hence, independent term is
$$^{10}{C_5}\,{\sin ^5}P{\cos ^5}P = {\,^{10}}{C_5}\frac{{{{\sin }^5}2p}}{{32}}$$
which is greatest when $$\sin 2p = 1.$$

$$\because \,\,{x^3}$$ and higher powers of $$x$$ may be neglected
$$\eqalign{ & \therefore \,\,\frac{{{{\left( {1 + x} \right)}^{\frac{3}{2}}} - {{\left( {1 + \frac{x}{2}} \right)}^3}}}{{{{\left( {1 - x} \right)}^{\frac{1}{2}}}}} \cr & = {\left( {1 - x} \right)^{\frac{{ - 1}}{2}}}\left[ {\left( {1 + \frac{3}{2}x + \frac{{\frac{3}{2}.\frac{1}{2}}}{{2!}}{x^2}} \right) - \left( {1 + \frac{{3x}}{2} + \frac{{3.2}}{{2!}}\frac{{{x^2}}}{4}} \right)} \right] \cr & = \left[ {1 + \frac{x}{2} + \frac{{\frac{1}{2}.\frac{3}{2}}}{{2!}}{x^2}} \right]\left[ {\frac{{ - 3}}{8}{x^2}} \right] = \frac{{ - 3}}{8}{x^2} \cr} $$
(as $${x^3}$$ and higher powers of $$x$$ can be neglected)

Putting the value of $${C_0},{C_2},{C_4}.....,$$   we get
$$\eqalign{ & = 1 + \frac{{n\left( {n - 1} \right)}}{{3 \cdot 2!}} + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{5 \cdot 4!}} + ..... \cr & = \frac{1}{{n + 1}}\left[ {\left( {n + 1} \right) + \frac{{\left( {n + 1} \right)n\left( {n - 1} \right)}}{{3!}} + \frac{{\left( {n + 1} \right)n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{5!}} + .....} \right] \cr & {\text{Put, }}n + 1 = N \cr & = \frac{1}{N}\left[ {N + \frac{{N\left( {N - 1} \right)\left( {N - 2} \right)}}{{3!}} + \frac{{N\left( {N - 1} \right)\left( {N - 2} \right)\left( {N - 3} \right)\left( {N - 4} \right)}}{{5!}} + .....} \right] \cr & = \frac{1}{N}\left\{ {^N{C_1} + {\,^N}{C_3} + {\,^N}{C_5} + .....} \right\} \cr & = \frac{1}{N}\left\{ {{2^{N - 1}}} \right\} = \frac{{{2^n}}}{{n + 1}}\,\,\,\,\left\{ {\because N = n + 1} \right\} \cr} $$

General term of the given series is
$$r\frac{{^n{C_r}}}{{^n{C_{r - 1}}}} = n + 1 - r$$
By taking summation over $$n,$$ we get
$$\eqalign{ & \sum\limits_1^{15} {r\frac{{^n{C_r}}}{{^n{C_{r - 1}}}}} = \sum\limits_{n = 1}^{15} {\left( {n + 1 - r} \right)} = \sum\limits_1^{15} {\left( {16 - r} \right)} \cr & = 16 \times 15 - \frac{1}{2} \cdot 15 \times 16 \cr} $$
By using sum of $$n$$ natural numbers $$ = \frac{{n\left( {n + 1} \right)}}{2}$$
$$= 240 - 120 = 120$$

Given that $$r$$ and $$n$$ are +ve integers such that $$r > 1, n > 2$$
Also in the expansion of $${\left( {1 + x} \right)^{2n}}$$
co - eff. of $${\left( {3r} \right)^{th}}$$  term = co - eff. of $${\left( {r + 2} \right)^{th}}$$  term
$$\eqalign{ & \Rightarrow \,{\,^{2n}}{C_{3r - 1}} = {\,^{2n}}{C_{r + 1}} \cr & \Rightarrow \,\,3r - 1 = r + 1\,\,{\text{or }}3r - 1 + r + 1 = 2n \cr & \Rightarrow \,\,r = 1\,\,{\text{or }}2r = n \cr & {\text{But }}\,r > 1 \cr & \therefore \,\,n = 2r \cr} $$

$$\frac{{r \cdot {\,^n}{C_r}}}{{^n{C_{r - 1}}}} = \frac{{n \cdot {\,^{n - 1}}{C_{r - 1}}}}{{^n{C_{r - 1}}}} = n \cdot \frac{{\left( {n - 1} \right)!}}{{\left( {r - 1} \right)!\left( {n - r} \right)!}} \times \frac{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}{{n!}} = n - r + 1.$$
∴ sum $$ = n + \left( {n - 1} \right) + ..... + \left( {n - 9} \right).$$

The given sigma is the expansion of $$\left[ {\left( {x - 3} \right) + 2} \right]100 = \left( {x - 1} \right)100 = \left( {1 - x} \right)100$$
Therefore, $$x^{53}$$ will occur in $$T_{54} .$$
$${T_{54}} = {\,^{100}}{C_{53}}{\left( { - x} \right)^{53}}$$
Therefore, the coefficient is $$ - {\,^{100}}{C_{53}}$$

$$\eqalign{ & ^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}{C_3}} \cr & \Rightarrow \,{\,^{50}}{C_4} + \left[ {^{55}{C_3} + {\,^{54}}{C_3} + {\,^{53}}{C_3} + {\,^{52}}{C_3} + {\,^{51}}{C_3} + {\,^{50}}{C_3}} \right] \cr & {\text{We know }}\left[ {^n{C_r} + {\,^n}{C_{r - 1}} = {\,^{n + 1}}{C_r}} \right] \cr & \Rightarrow \,\,\left( {^{50}{C_4} + {\,^{50}}{C_3}} \right) + {\,^{51}}{C_3} + {\,^{52}}{C_3} + {\,^{53}}{C_3} + {\,^{54}}{C_3} + {\,^{55}}{C_3} \cr & \Rightarrow \,\,\left( {^{51}{C_4} + {\,^{51}}{C_3}} \right) + {\,^{52}}{C_3} + {\,^{53}}{C_3} + {\,^{54}}{C_3} + {\,^{55}}{C_3} \cr} $$
Proceeding in the same way, we get
$$ \Rightarrow \,{\,^{55}}{C_4} + {\,^{55}}{C_3} = {\,^{56}}{C_4}.$$

Given expression can be written as
$$\eqalign{ & {\left( {\left( {{x^{\frac{1}{3}}} + 1} \right) - \left( {\frac{{\sqrt x + 1}}{{\sqrt x }}} \right)} \right)^{10}} \cr & = {\left( {{x^{\frac{1}{3}}} + 1 - 1 - \frac{1}{{\sqrt x }}} \right)^{10}} \cr & = {\left( {{x^{\frac{1}{3}}} - {x^{ - \frac{1}{2}}}} \right)^{10}} \cr} $$
General term $$ = {T_{r + 1}} = {\,^{10}}{C_r}{\left( {{x^{\frac{1}{3}}}} \right)^{10 - r}}{\left( { - {x^{ - \frac{1}{2}}}} \right)^r}$$
$$ = {\,^{10}}{C_r}{x^{\frac{{10 - r}}{3}}}.{\left( { - 1} \right)^r}.{x^{ - \frac{r}{2}}} = {\,^{10}}{C_r}{\left( { - 1} \right)^r}.{x^{\frac{{10 - r}}{3}.\frac{r}{2}}}$$
Term will be independent of $$x$$ when $$\frac{{10 - r}}{3} - \frac{r}{2} = 0$$
$$ \Rightarrow \,\,r = 4$$
So, required term $$ = {T_5} = {\,^{10}}{C_4} = 210$$

We have $$\left( {^{21}{C_1} + {\,^{21}}{C_2}...... + {\,^{21}}{C_{10}}} \right) - \left( {^{10}{C_1} + {\,^{10}}{C_2}...... + {\,^{10}}{C_{10}}} \right)$$
$$\eqalign{ & = \frac{1}{2}\left[ {\left( {^{21}{C_1} + ...... + {\,^{21}}{C_{10}}} \right) + \left( {^{21}{C_{11}} + ......{\,^{21}}{C_{20}}} \right)} \right] - \left( {{2^{10}} - 1} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\left( {\because \,{\,^{10}}{C_1} + {\,^{10}}{C_2} + ...... + {\,^{10}}{C_{10}} = {2^{10}} - 1} \right) \cr & = \frac{1}{2}\left[ {{2^{21}} - 2} \right] - \left( {{2^{10}} - 1} \right) \cr & = \left( {{2^{20}} - 1} \right) - \left( {{2^{10}} - 1} \right) = {2^{20}} - {2^{10}} \cr} $$