Binomial Theorem MCQ Questions & Answers in Algebra | Maths

$${\left\{ {1 + \left( {{x^2} + \frac{1}{{{x^2}}}} \right)} \right\}^n} = {\,^n}{C_0} + {\,^n}{C_1}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) + {\,^n}{C_2}{\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^2} + ..... + {\,^n}{C_n}{\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^n}$$
Here, all the terms are positive and will contain powers
$${\left( {{x^2}} \right)^0},{\left( {{x^2}} \right)^1},{\left( {{x^2}} \right)^2},.....,{\left( {{x^2}} \right)^n},{\left( {{x^2}} \right)^{ - n}},{\left( {{x^2}} \right)^{ - \left( {n - 1} \right)}},.....,{\left( {{x^2}} \right)^{ - 1}}\,{\text{only}}{\text{.}}$$
∴ the number of terms will be $$2n + 1.$$

$$\eqalign{ & {\left( {1073} \right)^{71}} - m = {\left( {73 + 1000} \right)^{71}} - m \cr & = {\,^{71}}{C_0}{\left( {73} \right)^{71}} + {\,^{71}}{C_1}{\left( {73} \right)^{70}}\left( {1000} \right) + {\,^{71}}{C_2}{\left( {73} \right)^{69}}{\left( {1000} \right)^2} + ..... + {\,^{71}}{C_{71}}{\left( {1000} \right)^{71}} - m \cr} $$
Above will be divisible by 10 if $$^{71}{C_0}{\left( {73} \right)^{71}}$$   is divisible by 10
Now, $$^{71}{C_0}{\left( {73} \right)^{71}} = {\left( {73} \right)^{70}} \cdot 73 = {\left( {{{73}^2}} \right)^{35}} \cdot 73$$
The last digit of $$73^2$$ is 9, so the last digit of $${\left( {{{73}^2}} \right)^{35}}$$  is 9.
$$\therefore $$ Last digit of $${\left( {{{73}^2}} \right)^{35}} \cdot 73{\text{ is }}7$$
Hence, the minimum positive integral value of $$m$$ is 7, so that it is divisible by 10.

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

$$\eqalign{ & {\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}} \cr & = {\left[ {{{\left( {\sqrt 3 + 1} \right)}^2}} \right]^n} - {\left[ {{{\left( {\sqrt 3 - 1} \right)}^2}} \right]^n} \cr & = {\left( {4 + 2\sqrt 3 } \right)^n} - {\left( {4 - 2\sqrt 3 } \right)^n} \cr & = {2^n}\left[ {{{\left( {2 + \sqrt 3 } \right)}^n} - {{\left( {2 - \sqrt 3 } \right)}^n}} \right] \cr & = {2^n} \times 2\left[ {^n{C_1}{2^{n - 1}}\sqrt 3 + {\,^n}{C_3}{{.2}^{n - 3}}3\sqrt 3 + .....} \right] \cr & = {2^{n + 1}}\sqrt 3 \left[ {^n{C_1}{{.2}^{n - 1}} + {\,^n}{C_3}{2^{n - 3}}.3 + .....} \right] \cr & = \sqrt 3 \times \,{\text{Some integer}} \cr & \therefore \,\,{\text{irrational number}} \cr} $$

We have,
$$\eqalign{ & {7^9} + {9^7} = {\left( {8 - 1} \right)^9} + {\left( {8 + 1} \right)^7} = {\left( {1 + 8} \right)^7} - {\left( {1 - 8} \right)^9} \cr & = \left[ {1 + {\,^7}{C_1} 8 + {\,^7}{C_2} {8^2} + ..... + {\,^7}{C_7} {8^7}} \right] - \left[ {1 - {\,^9}{C_1} 8 + {\,^9}{C_2} {8^2} - ..... - {\,^9}{C_9} {8^9}} \right] \cr & = {\,^7}{C_1} 8 + {\,^9}{C_1} 8 + \left[ {^7{C_2} + {\,^7}{C_3} 8 + ..... - {\,^9}{C_2} + {\,^9}{C_3} 8 - .....} \right]{8^2} \cr & = 8\left( {7 + 9} \right) + 64k = 8 \cdot 16 + 64k = 64q, \cr} $$
where $$q = k + 2$$
Thus, $$7^9 + 9^7$$  is divisible by 64.

Here, $${T_{r + 1}} = {\,^{10}}{C_r}{\left( {\sqrt 2 } \right)^{10 - r}}{\left( {{3^{\frac{1}{5}}}} \right)^r},$$      where $$r = 0, 1, 2, ..... ,10.$$
We observe that in general term $$T_{r + 1}$$  powers of 2 and 3 are $$\frac{1}{2}\left( {10 - r} \right)$$   and $$\frac{1}{5}r$$  respectively and $$0 \leqslant r \leqslant 10.$$
So, both these powers will be integers together only when $$r = 0$$  or 10
$$\therefore $$ Sum of required terms $$ = {T_1} + {T_{11}}$$
$$ = {\,^{10}}{C_0}{\left( {\sqrt 2 } \right)^{10}} + {\,^{10}}{C_{10}}{\left( {{3^{\frac{1}{5}}}} \right)^{10}} = 32 + 9 = 41$$

We know that, $${\left( {1 + x} \right)^{20}} = {\,^{20}}{C_0} - {\,^{20}}{C_1}x + {\,^{20}}{C_2}{x^2} + ..... + {\,^{20}}{C_{10}}{x^{10}} + .....{\,^{20}}{C_{20}}{x^{20}}$$
$$\eqalign{ & {\text{Put }}x = - 1,\,\,\left( 0 \right) = {\,^{20}}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... + {\,^{20}}{C_{10}} - {\,^{20}}{C_{11}}..... + {\,^{20}}{C_{20}} \cr & \Rightarrow \,\,0 = 2\left[ {^{20}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... - {\,^{20}}{C_9} + {\,^{20}}{C_{10}}} \right] \cr & \Rightarrow \,{\,^{20}}{C_{10}} = 2\left[ {^{20}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... - {\,^{20}}{C_9} + {\,^{20}}{C_{10}}} \right] \cr & \Rightarrow \,{\,^{20}}{C_0} - {\,^{20}}{C_1} + {\,^{20}}{C_2} - {\,^{20}}{C_3} + ..... + {\,^{20}}{C_{10}} = \frac{1}{2}{\,^{20}}{C_{10}} \cr} $$

$$\eqalign{ & {\left( {1 + a - b + c} \right)^9} \cr & = \sum {\frac{{9!}}{{{x_1}!{x_2}!{x_3}!{x_4}!}} \cdot {{\left( 1 \right)}^{{x_1}}}{{\left( a \right)}^{{x_2}}}{{\left( { - b} \right)}^{{x_3}}}{{\left( c \right)}^{{x_4}}}} \cr} $$
$$ \Rightarrow $$ Coefficient of $${a^3}{b^4}c = \frac{{9!}}{{1!3!4!1!}} = \frac{{9!}}{{3!4!}}$$

$$\sum\limits_{i = 0}^m {^{10}{C_i}^{20}{C_{m - i}} = {\,^{10}}{C_0}{\,^{20}}{C_m} + {\,^{10}}{C_1}{\,^{20}}{C_{m - 1}} + {\,^{10}}{C_2}{\,^{20}}{C_{m - 2}}} + ..... + {\,^{10}}{C_m}{\,^{20}}{C_0}$$
= Co-eff of $${x^m}$$ in the expansion of product $${\left( {1 + x} \right)^{10}}{\left( {1 + x} \right)^{20}}$$
= Co-eff. of $${x^m}$$ in the expansion of $${\left( {1 + x} \right)^{30}}$$
$$ = \,{\,^{30}}{C_m}$$
To get max. value of given sum, $$^{30}{C_m}\,$$ should be max.
which is so when $$m = \frac{{30}}{2} = 15.$$
\[\left[ {{\rm{Using\, the\, fact\, that\, max }}\left( {^n{C_r}} \right) = \left\{ \begin{array}{l} ^n{C_{\frac{n}{2}}}\,\,{\rm{if }}\,n\,{\rm{is\, even}}\\ ^n{C_{\frac{{n + 1}}{2}}}\,{\rm{if }}\,n\,{\rm{is\, odd}} \end{array} \right.} \right]\]

We have
$$\eqalign{ & \sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{\,^n}{C_r}{x^r} = } \sum\limits_{r = 0}^n {r.{\,^n}{C_r}{x^r} + \sum\limits_{r = 0}^n {^n{C_r}{x^r}} } \cr & = \sum\limits_{r = 1}^n {r.\,\frac{n}{r}{\,^{n - 1}}{C_{r - 1}}{x^r} + {{\left( {1 + x} \right)}^n}} \cr & = nx\sum\limits_{r = 1}^n {^{n - 1}{C_{r - 1}}{x^{r - 1}} + {{\left( {1 + x} \right)}^n}} \cr & = nx{\left( {1 + x} \right)^{n - 1}} + {\left( {1 + x} \right)^n} = {\text{RHS}} \cr} $$
∴ Statement - 2 is correct.
Putting $$x = 1,$$  we get
$$\sum\limits_{r = 0}^n {\left( {r + 1} \right)\,{\,^n}{C_r} = n.\,{2^{n - 1}}} + {2^n} = \left( {n + 2} \right){.2^{n - 1}}.$$
∴ Statement - 1 is also true and statement - 2 is a correct explanation for statement - 1.