Binomial Theorem MCQ Questions & Answers in Algebra | Maths

$$\eqalign{ & {27^{40}} = {3^{120}} \cr & {3^{119}} = {\left( {4 - 1} \right)^{119}} = {\,^{119}}{C_0}{4^{119}} - {\,^{119}}{C_1}{4^{118}} + {\,^{119}}{C_2}{4^{117}} - {\,^{119}}{C_3}{4^{116}} + ..... + \left( { - 1} \right) \cr & \therefore {3^{119}} = 4k - 1 \cr & \therefore {3^{120}} = 12k - 3 = 12\left( {k - 1} \right) + 9 \cr} $$
$$\therefore $$ The required remainder is 9.

$$\eqalign{ & {S_n} = {a_0}C_0^2 + {a_1}C_1^2 + {a_2}C_2^2 + ..... + {a_n}C_n^2 \cr & {S_n} = {a_n}C_n^2 + {a_{n - 1}}C_{n - 1}^2 + {a_{n - 2}}C_{n - 2}^2 + ..... + {a_n}C_0^2 \cr & \overline {2{S_n} = \left( {{a_0} + {a_n}} \right)C_0^2 + \left( {{a_1} + {a_{n - 1}}} \right)C_1^2 + ..... + \left( {{a_n} + {a_0}} \right)C_n^1} \cr & 2{S_n} = \left( {2n + 2} \right)\left( {C_0^2 + C_1^2 + C_2^2 + ..... + C_n^2} \right) \cr & \therefore \,{S_n} = \left( {n + 1} \right){}^{2n}{C_n} \cr & \left[ {\because \,{a_0} + {a_n} = {a_1} + {a_{n - 1}} + ..... + = 2n + 2} \right] \cr} $$

Since we know that,
$$\eqalign{ & {\left( {x + a} \right)^5} + {\left( {x - a} \right)^5} \cr & = 2\left[ {^5{C_0}{x^5} + {\,^5}{C_2}{x^3} \cdot {a^2} + {\,^5}{C_4}x \cdot {a^4}} \right] \cr & \therefore \,\,{\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5} \cr & = 2\left[ {^5{C_0}{x^5} + {\,^5}{C_2}{x^3}\left( {{x^3} - 1} \right) + {\,^5}{C_4}x{{\left( {{x^3} - 1} \right)}^2}} \right] \cr & = 2\left[ {{x^5} + 10{x^6} - 10{x^3} + 5{x^7} - 10{x^4} + 5x} \right] \cr} $$
∴ Sum of co-efficients of odd degree terms = 2.

$${T_{r + 1}} = {\,^{256}}{C_r}{\left( {\sqrt 3 } \right)^{256 - r}}{\left( {\root 8 \of 5 } \right)^r} = {\,^{256}}{C_r}{\left( 3 \right)^{\frac{{256 - r}}{2}}}{\left( 5 \right)^{\frac{r}{8}}}$$
Terms will be integral if $$\frac{{256 - r}}{2}\,\& \,\frac{r}{8}$$   both are $$+ve$$  integer, which is so if $$r$$ is an integral multiple of 8. As $$0 \leqslant r \leqslant 256$$
$$\therefore r = 0,8,16,24,.....256,{\text{ total }}33{\text{ values}}{\text{.}}$$

$$\eqalign{ & {\text{Here, }}{P_n} = {\,^n}{C_0} \cdot {\,^n}{C_1} \cdot {\,^n}{C_2}.....{\,^n}{C_n} \cr & {\text{and }}{P_{n + 1}} = {\,^{n + 1}}{C_0} \cdot {\,^{n + 1}}{C_1} \cdot {\,^{n + 1}}{C_2}.....{\,^{n + 1}}{C_{n + 1}} \cr & \therefore \frac{{{P_{n + 1}}}}{{{P_n}}} = \frac{{^{n + 1}{C_0} \cdot {\,^{n + 1}}{C_1} \cdot {\,^{n + 2}}{C_2}.....{\,^{n + 1}}{C_{n + 1}}}}{{^n{C_0} \cdot {\,^n}{C_1} \cdot {\,^n}{C_2}.....{\,^n}{C_n}}} \cr & = \left( {\frac{{^{n + 1}{C_1}}}{{^n{C_0}}}} \right)\left( {\frac{{^{n + 1}{C_2}}}{{^n{C_1}}}} \right)\left( {\frac{{^{n + 1}{C_3}}}{{^n{C_0}}}} \right).....\left( {\frac{{^{n + 1}{C_{n + 1}}}}{{^n{C_n}}}} \right) \cr & = \left( {\frac{{n + 1}}{1}} \right)\left( {\frac{{n + 1}}{2}} \right)\left( {\frac{{n + 1}}{3}} \right).....\left( {\frac{{n + 1}}{{n + 1}}} \right) \cr & = \frac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}} \cr} $$

$$\eqalign{ & {T_n} = \frac{{1 \cdot 3 \cdot 5.....\left( {2n - 1} \right)}}{{\left( {2n} \right)!}} \cr & = \frac{{1 \cdot 2 \cdot 3 \cdot 4.....\left( {2n - 1} \right) \cdot 2n}}{{\left( {2n} \right)!\,2 \cdot 4 \cdot 6.....2n}} \cr & = \frac{{\left( {2n} \right)!}}{{\left( {2n} \right)!\,{2^n}n!}} = \frac{1}{{{2^n}n!}} \cr} $$
Now putting $$n = 1, 2, 3, .....$$    we see that the sum of series
$$\eqalign{ & S = \frac{1}{2} + \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( {\frac{1}{2}} \right)}^3}}}{{3!}} + ..... \cr & = {e^{\frac{1}{2}}} - 1 = \sqrt e - 1 \cr} $$

Let the consecutive coefficient of $${\left( {1 + x} \right)^n}$$  are $$^n{C_{r - 1}},{\,^n}{C_r},{\,^n}{C_{r + 1}}$$
From the given condition, $$^n{C_{r - 1}}:{\,^n}{C_r}:{\,^n}{C_{r + 1}} = 6:33:110$$
Now, $$^n{C_{r - 1}}:{\,^n}{C_r}= 6:33$$
$$\eqalign{ & \Rightarrow \,\frac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}} \times \frac{{r!\left( {n - r} \right)!}}{{n!}} = \frac{6}{{33}} \cr & \Rightarrow \,\frac{r}{{n - r + 1}} = \frac{2}{{11}} \cr & \Rightarrow \,11r = 2n - 2r + 2 \cr & \Rightarrow \,2n - 13r + 2 = 0\,\,\,.....\left( {\text{i}} \right) \cr & {\text{and}}{{\text{ }}^n}{C_r}:{\,^n}{C_{r + 1}} = 33\,:\,110 \cr & \Rightarrow \,\frac{{n!}}{{r!\left( {n - r} \right)!}} \times \frac{{\left( {r + 1} \right)!\left( {n - r - 1} \right)!}}{{n!}} = \frac{{33}}{{110}} = \frac{3}{{10}} \cr & \Rightarrow \,\frac{{\left( {r + 1} \right)}}{{n - r}} = \frac{3}{{10}} \cr & \Rightarrow \,3n - 13r - 10 = 0\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Solving (i) & (ii), we get $$n = 12$$

The middle term in the expansion of
$${\left( {1 + \alpha x} \right)^4} = {T_3} = {\,^4}{C_2}{\left( {\alpha x} \right)^2} = 6{\alpha ^2}{x^2}$$
The middle term in the expansion of
$${\left( {1 - \alpha x} \right)^6} = {T_4} = {\,^6}{C_3}{\left( { - \alpha x} \right)^3} = - 20{\alpha ^3}{x^3}$$
According to the question
$$\eqalign{ & 6{\alpha ^2} = - 20{\alpha ^3} \cr & \Rightarrow \,\,\alpha = - \frac{3}{{10}} \cr} $$

$$\eqalign{ & {\left( {1 - x - {x^2} + {x^3}} \right)^6} = {\left[ {\left( {1 - x} \right) - {x^2}\left( {1 - x} \right)} \right]^6} \cr & = {\left( {1 - x} \right)^6}{\left( {1 - {x^2}} \right)^6} \cr & = \left( {1 - 6x + 15{x^2} - 20{x^3} + 15{x^4} - 6{x^5} + {x^6}} \right) \times \left( {1 - 6{x^2} + 15{x^4} - 20{x^6} + 15{x^8} - 6{x^{10}} + {x^{12}}} \right) \cr & {\text{Co - efficient of }}{x^7} = \left( { - 6} \right)\left( { - 20} \right) + \left( { - 20} \right)\left( {15} \right) + \left( { - 6} \right)\left( { - 6} \right) \cr & = - 144 \cr} $$

$${t_{r + 1}} = {\,^{100}}{C_r}{\left( {\root 8 \of 5 } \right)^{100 - r}} \cdot {\left( {\root 6 \of 2 } \right)^r}.$$
As 2 and 5 are coprime, $${t_{r + 1}}$$ will be rational if $$100 - r$$   is a multiple of 8 and $$r$$ is a multiple of 6. Also $$0 \leqslant r \leqslant 100.$$
$$\eqalign{ & \therefore \,\,r = 0,6,12,.....,96 \cr & \therefore \,\,100 - r = 4,10,16,.....,100\,\,\,\,\,\,.....\left( 1 \right) \cr} $$
But $$100 - r$$   is to be a multiple of 8.
So, $$100 - r = 0,8,16,24,.....,96\,\,\,\,\,\,\,\,.....\left( 2 \right)$$
The common terms in (1) and (2) are 16, 40, 64 and 88.
∴ $$r = 84, 60, 36, 12$$     give rational terms.
∴ the number of irrational terms = $$101 - 4 = 97.$$