CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 11 - MCQExams.com

There are total of $$n$$ brackets. The term $$x^{n-2}$$ will be formed when integers are chosen from any two brackets and $$x$$ is chosen from all the other brackets and multiplied.

Thus, the Coefficient of $$x^{n-2}$$ is 

$$ C= (1\times2+1\times3+...+1\times n)+(2\times3+2\times4+..+2\times n)+...+((n-1)\times n)$$

$$ = \left(\dfrac {(n)(n+1)}{2}-1\right)+2\left(\dfrac {(n)(n+1)}{2}-1-2\right)+...+(n-1)\left(\dfrac {(n)(n+1)}{2}-(1+2+3+..+(n-1))\right) $$

$$ = \left\{(1+2+3+...+(n-1))(\dfrac {(n)(n+1)}{2})\right\} - \left\{ 1+2(1+2)+3(1+2+3)+...+(n-1)(1+...+n-1)\right\}$$

$$ = \left\{(\dfrac {(n-1)(n)}{2})(\dfrac {(n)(n+1)}{2})\right\} - \left\{\sum_{1}^{n-1} k(\dfrac {(k)(k+1)}{2})\right\}$$

$$= \left\{ \dfrac {n^{2}(n^{2}-1)}{4}\right\} - \left\{ \sum_{1}^{n-1} \dfrac {k^{3}+k^{2}}{2} \right\} $$

$$ = \left\{ \dfrac {n^{2}(n^{2}-1)}{4}\right\} - \dfrac {1}{2}\left\{ (\dfrac {(n-1)(n)}{2})^{2} + \dfrac {(n-1)n(2n-1)}{6} \right\} $$

$$ = \dfrac {n(n-1)}{4} \left\{ {n(n+1)} -  \dfrac {n(n-1)}{2} - \dfrac {2n-1}{3} \right\} $$

$$ = \dfrac {n(n-1)}{4}\left\{ \dfrac {n(n+3)}{2} -\dfrac {2n-1}{3} \right\} $$

$$ = \dfrac {n(n-1)}{4}\left\{ \dfrac {3n^{2}+9n-4n+2}{6} \right\} $$

$$ = \dfrac {n(n-1)}{4}\left\{ \dfrac {(3n+2)(n+1)}{6} \right\} $$

$$ \therefore $$ Option $$B$$ is correct.

Put $$y={ 3 }^{ \log _{ 3 }{ \sqrt { { 9 }^{ \left| x-2 \right|  } }  }  }$$ $$\Rightarrow$$ $$\log _{ 3 }{ y } =\log _{ 3 }{ \sqrt { { 9 }^{ \left| x-2 \right|  } }  } $$
$$\Rightarrow \quad y=\sqrt { { 9 }^{ \left| x-2 \right|  } } ={ 3 }^{ \left| x-2 \right|  }$$
Now, put $$z={ 7 }^{ (\tfrac 15)\log _{ 7 }{ \left[ (4).{ 3 }^{ \left| x-2 \right|  }-9 \right]  }  }$$
$$\log _{ 7 }{ z } =\cfrac { 1 }{ 5 } \log _{ 7 }{ \left[ (4).{ 3 }^{ \left| x-2 \right|  }-9 \right]  } $$
$$={ \left( \log _{ 7 }{ \left[ (4).{ 3 }^{ \left| x-2 \right|  }-9 \right]  }  \right)  }^{ \tfrac 15 }$$
$$\Rightarrow z={ \left[ (4).{ 3 }^{ \left| x-2 \right|  }-9 \right]  }^{ \tfrac 15 }$$
Now, $$E={ \left( y+z \right)  }^{ 7 }$$ and 6th term is given by
$${ t }_{ 6 }={ _{  }^{ 7 }{ C } }_{ 5 }{ y }^{ 7-5 }{ z }^{ 5 }=21{ \left( { 3 }^{ \left| x-2 \right|  } \right)  }^{ 2 }\left\{ (4).{ 3 }^{ \left| x-2 \right|  }-9 \right\} $$
$$\Rightarrow 567=21{ \left( { 3 }^{ 2\left| x-2 \right|  } \right)  }\left\{ (4).{ 3 }^{ \left| x-2 \right|  }-9 \right\} $$
$$\Rightarrow 27=\left[ (4){ 3 }^{ 3\left| x-2 \right|  } \right] -\left[ (9)({ 3 }^{ 2\left| x-2 \right|  }) \right] $$
$$\Rightarrow \quad 27=(4){ 3 }^{ 3\left| x-2 \right|  }-(9){ 3 }^{ 2\left| x-2 \right|  }$$
$$\Rightarrow \quad 4{ u }^{ 3 }-9{ u }^{ 2 }-27=0\quad ;\quad u={ 3 }^{ \left| x-2 \right|  }$$
note that $$u=3$$ satisfies this equation
$${ 3 }^{ \left| x-2 \right|  }=3\quad \Rightarrow \left| x-2 \right| =1\quad \Rightarrow \quad x-2=\pm 1\quad $$
$$x=2\pm1=3\quad or\quad 1$$.

$${ (1+x) }^{ n }=^{ n }C_{ 0 }+^{ n }C_{ 1 }x+^{ n }C_{ 2 }{ x }^{ 2 }.........^{ n }C_{ n }{ x }^{ n }\\ \Rightarrow ^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }.........^{ n }C_{ n }={ 2 }^{ n }$$ 

For the given expansion $${ (1+x) }^{ 59 }$$ let sum of last $$30$$ coefficients be $$S$$

$$S=^{ 59 }C_{ 30 }+^{ 59 }C_{ 31 }+^{ 59 }C_{ 32 }..........^{ 59 }C_{ 59 }$$    ....(i)

As $$^{ n }C_{ r }=^{ n }C_{ n-r }$$

$$\Rightarrow S=^{ 59 }C_{ 29 }+^{ 59 }C_{ 28 }+^{ 59 }C_{ 27 }..........^{ 59 }C_{ 0 }$$    ....(ii)

Adding (i) and (ii), we get

$$\Rightarrow 2S=^{ 59 }C_{ 0 }+^{ 59 }C_{ 1 }+^{ 59 }C_{ 2 }........^{ 59 }C_{ 59 }\\ \Rightarrow 2S={ 2 }^{ 59 }\\ \Rightarrow S={ 2 }^{ 58 }$$

So, option B is correct.

Consider the given expression.

$$\Rightarrow {{\left( {{2}^{x}}+\dfrac{1}{{{4}^{x}}} \right)}^{n}}$$

Given:

$$\dfrac{{{t}_{3}}}{{{t}_{2}}}=7$$

By binomial expansion, we know that

$$ {{t}_{3}}={}^{n}{{C}_{2}}{{\left( {{2}^{x}} \right)}^{n-2}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{2}} $$

$$ {{t}_{2}}={}^{n}{{C}_{1}}{{\left( {{2}^{x}} \right)}^{n-1}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{1}} $$

Again, we have

$$ {}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}=36 $$

$$ {}^{n+1}{{C}_{2}}=36 $$

$$ \dfrac{\left( n+1 \right)!}{2!\left( n+1-2 \right)!}=36 $$

$$ \dfrac{n\left( n+1 \right)}{2}=36 $$

$$ {{n}^{2}}+n-72=0 $$

$$ n=8,-9 $$

Therefore,

$$n=8$$

Now,

$$ \dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{}^{n}{{C}_{2}}{{\left( {{2}^{x}} \right)}^{n-2}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{2}}}{{}^{n}{{C}_{1}}{{\left( {{2}^{x}} \right)}^{n-1}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{1}}} $$

$$ \dfrac{{}^{8}{{C}_{2}}{{\left( {{2}^{x}} \right)}^{8-2}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{2}}}{{}^{8}{{C}_{1}}{{\left( {{2}^{x}} \right)}^{8-1}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{1}}}=7 $$

$$ \dfrac{28{{\left( {{2}^{x}} \right)}^{6}}{{\left( \dfrac{1}{{{\left( {{2}^{2}} \right)}^{x}}} \right)}^{2}}}{8{{\left( {{2}^{x}} \right)}^{7}}{{\left( \dfrac{1}{{{\left( {{2}^{2}} \right)}^{x}}} \right)}^{1}}}=7 $$

$$ \dfrac{7\times {{2}^{6x-4x}}}{2\times {{2}^{7x-2x}}}=7 $$

$$ {{2}^{2x-5x}}=2 $$

$$ -3x=1 $$

$$ x=-\dfrac{1}{3} $$

Hence, this is the required result.