Binomial Theorem MCQ Questions & Answers in Algebra | Maths

Given expansion is $${\left( {\sqrt x + \frac{k}{{{x^2}}}} \right)^{10}}$$
$$\eqalign{ & {\left( {r + 1} \right)_{th}}\,{\text{term}},\,{T_{r + 1}} = {\,^{10}}{C_r}{\left( {\sqrt x } \right)^{10 - r}}{\left( {\frac{k}{{{x^2}}}} \right)^r} \cr & \Rightarrow \,{T_{r + 1}} = {\,^{10}}{C_r}{x^{\frac{{5 - r}}{2}}} \cdot {\left( k \right)^r} \cdot {x^{ - 2r}} \cr & \therefore \,{T_{r + 1}} = {\,^{10}}{C_r}{x^{\frac{{\left( {10 - 5r} \right)}}{2}}}{\left( k \right)^r} \cr} $$
Since, $${T_{r + 1}}$$  is independent of $$x$$
$$\eqalign{ & \therefore \,\frac{{10 - 5r}}{2} = 0 \cr & \Rightarrow \,r = 2 \cr & \therefore 405 = {\,^{10}}{C_2}{\left( k \right)^2} \cr & 405 = 45 \times {k^2} \cr & \Rightarrow \,{k^2} = 9 \cr & \Rightarrow \,k = \pm \,3 \cr} $$

$$\eqalign{ & A = \sum\limits_{n = 1}^\infty {\frac{{2n - 1 + 1}}{{\left( {2n - 1} \right)!}}} \cr & = \sum\limits_{n = 1}^\infty {\left[ {\frac{1}{{\left( {2n - 2} \right)!}} + \frac{1}{{\left( {2n - 1} \right)!}}} \right]} \cr & = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + ..... = e \cr} $$
Similarly $$B = e^{- 1}$$  as terms will be alternately positive and negative.
$$\therefore AB = e \cdot {e^{ - 1}} = 1$$

In $${\left( {1 - x} \right)^n}$$  the numerically greatest co-efficient is the middle co-efficient.
From the question the middle term is the $${5^{th}}$$ term.
∴ there will be 9 terms. So, $$n = 8.$$

As the index = 20, the $${11^{th}}$$ term is the middle term. So, we have to find the $${14^{th}}$$ term.

The required sum $$ = {\,^{19}}{C_{10}} + {\,^{19}}{C_{11}} + ..... + {\,^{19}}{C_{19}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\,^{19}}{C_0} + {\,^{19}}{C_1} + {\,^{19}}{C_2} + ..... + {\,^{19}}{C_9}\left( {\because \,{\,^n}{C_r} = {\,^n}{C_{n - r}}} \right)$$
Adding, $$2 \times $$  (required sum) $$ = {\,^{19}}{C_0} + {\,^{19}}{C_1} + ..... + {\,^{19}}{C_{19}} = {2^{19}}.$$

$$\eqalign{ & ^m{C_m} + {\,^{m + 1}}{C_m} + ..... + {\,^n}{C_m} \cr & = {\,^{m + 1}}{C_{m + 1}} + {\,^{m + 1}}{C_m} + ..... + {\,^n}{C_m} \cr & = {\,^{m + 2}}{C_{m + 1}} + {\,^{m + 2}}{C_m} + ..... + {\,^n}{C_m} \cr & .................................... \cr & = {\,^n}{C_{m + 1}} + {\,^{n}}{C_m} = {\,^{n + 1}}{C_{m + 1}} \cr} $$

On expansion and simplification,
expression $$ = 2\left\{ {^6{C_0}{x^6} + {\,^6}{C_2}{x^4}\left( {{x^2} - 1} \right) + {\,^6}{C_4}{x^2}{{\left( {{x^2} - 1} \right)}^2} + {\,^6}{C_6}{{\left( {{x^2} - 1} \right)}^3}} \right\}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left\{ {\left( {^6{C_0} + {\,^6}{C_2} + {\,^6}{C_4} + {\,^6}{C_6}} \right){x^6} + \left( {{ - ^6}{C_2} - {\,^6}{C_4} \times 2 - {\,^6}{C_6} \times 3} \right){x^4} + \left( {^6{C_4} + {\,^6}{C_6} \times 3} \right){x^2} - {\,^6}{C_6}} \right\}.$$

Let $${\left( {\sqrt 2 + 1} \right)^6} = k + f,$$    where $$k$$ is integral part and $$f$$ the fraction $$\left( {0 \leqslant f < 1} \right)$$
Let $${\left( {\sqrt 2 - 1} \right)^6} = f',\left( {0 < f' < 1} \right),$$
Since, $$0 < \left( {\sqrt 2 - 1} \right) < 1$$
Now, $$k + f + f' = {\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6}$$
$$\eqalign{ & = 2\left\{ {^6{C_0} \cdot {2^3} + {\,^6}{C_2} \cdot {2^2} + {\,^6}{C_4} \cdot 2 + {\,^6}{C_6}} \right\} = 198\,\,\,.....\left( {\text{i}} \right) \cr & \therefore f + f' = 198 - k = {\text{an integer}} \cr & {\text{But, }}0 \leqslant f < 1{\text{ and }}0 < f' < 1 \cr & \Rightarrow 0 < \left( {f + f'} \right) < 2 \cr & \Rightarrow f + f' = 1,\left( {\because f + f'{\text{ is an integer}}} \right) \cr & \therefore {\text{By}}\left( {\text{i}} \right),I = 198 - \left( {f + f'} \right) = 198 - 1 = 197 \cr} $$

Co-eff. of $${x^n}$$ in $$\left( {1 + x} \right){\left( {1 - x} \right)^n}$$
= Co-eff. of $${x^n}$$ in $${\left( {1 - x} \right)^n} + $$   Co-eff. of $${x^{n - 1}}$$ in $${\left( {1 - x} \right)^n}$$
$$\eqalign{ & = {\left( { - 1} \right)^n}{\,^n}{C_n} + {\left( { - 1} \right)^{n - 1}}{\,^n}{C_{n - 1}} = {\left( { - 1} \right)^n}1 + {\left( { - 1} \right)^{n - 1}}n \cr & = {\left( { - 1} \right)^n}\left( {1 - n} \right) \cr} $$