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Binomial Theorem Test - 60

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Binomial Theorem Test - 60
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  • Question 1
    1 / -0
    $$\sum { { \left( -1 \right)  }^{ r } } ~ { _{  }^{ n }{ C } }_{ r }\cfrac { 1+r\log _{ e }{ 10 }  }{ { \left( 1+\log _{ e }{ { 10 }^{ n } }  \right)  }^{ r } } $$
    Solution

  • Question 2
    1 / -0
    If $$\left\{ x \right\}$$  denotes the fraction part of $$'x'$$, then $$\left\{ \dfrac { { 3 }^{ 1001 } }{ 82 }  \right\} =$$
    Solution
    $$ \Rightarrow \left \{ \dfrac{3^{1001}}{82} \right \}$$
    $$ = 3^{4} \rightarrow 81 $$   $$ 1-\dfrac{1}{82} -1 $$
    $$ = \dfrac{(81)^{997}}{82} $$            $$ \boxed{\dfrac{81}{82}} $$
    $$ = \dfrac{(82-1)^{997}}{82}$$
    $$ = \left \{ -\dfrac{(1-82)^{997}}{82} \right \}$$
    $$ = ^{997}C_{0} (-82)^0 + ^{997}C_{1} \dfrac{(-82)^1}{82} $$
    $$ = -\dfrac{1}{82}$$
    All will in proper division of 82.
    $$ = 1-\dfrac{1}{82}-1 $$
    $$ = \left \{ \dfrac{81}{82}-1 \right \}$$
    $$ = \dfrac{81}{82}$$

  • Question 3
    1 / -0
    The coefficient of $$x^{160}$$ in the expansion of $$\displaystyle (x^8 + 1)^{60} \left( x^{12} + 3x^4 + \frac{3}{x^4} + \frac{1}{x^{12}} \right)^{-10}$$ is
    Solution
    $$S = { \left( 1+{ x }^{ 8 } \right)  }^{ 60 }.{ \left( { x }^{ 12 }+{ 3x }^{ 4 }+\dfrac { 3 }{ { x }^{ 4 } } +\dfrac { 1 }{ { x }^{ 12 } }  \right)  }^{ -10 }$$
        $$={ x }^{ 12 }+{ 3x }^{ 4 }+\dfrac { 3 }{ { x }^{ 4 } } +\dfrac { 1 }{ { x }^{ 12 } } ={ \left( { x }^{ 4 }+\dfrac { 1 }{ { x }^{ 4 } }  \right)  }^{ 3 }$$
    $$\Rightarrow 3 = { (1+{ x }^{ 8 }) }^{ 60 }.{ \left\{ { \left( \dfrac { { x }^{ 8 }+1 }{ { x }^{ 4 } }  \right)  }^{ 3 } \right\}  }^{ -10 }$$
              $$=\dfrac { { (1+{ x }^{ 8 }) }^{ 60 }.{ (1+{ x }^{ 8 }) }^{ -30 } }{ { x }^{ -120 } } $$

             $$={ (1+{ x }^{ 8 }) }^{ 30 }.{ x }^{ 120 }$$
             $$=[1+{ ^{ 30 }C }_{ 1 }.{ x }^{ 8 }+.....{ ^{ 30 }C }_{ 5 }.{ (x }^{ 8 })^{ 5 }].{ x }^{ 120 }$$
    $$\therefore$$ Coefficient of $${ x }^{ 160 }={ ^{ 30 }C }_{ 5 }.$$
    Hence the answer is $${ ^{ 30 }C }_{ 5 }.$$
  • Question 4
    1 / -0
    If $$C_r \, = \, (^{100 C _ r }) , then \,  E = \sum_{r = 0 }^{n + 4 } (-1)^r  \, c_r \, c_{r + 1}$$
    Solution
    $$(1 \, + \, x)^{101} \, = \, C_0 \, + \, C_1 x \, + \, C_2 \, x^2 \, + \, .....$$
    $$(1 \, - \, x)^{101} \, = \, C_0 \, x^{101} \, - C_1 \, x^{100} \, + \, C_2 \, x^{99} \, - \, ....$$
    Multiplying both sides, 
    $$(x^2 \, - \, 1)^{101} \, = \, (C_0 \, + \, C_1 \, x \, + \, C_2 \, x^2 \, + \, ....)$$
    $$(C_0 \, x^{101} \, - \, C_1 \, x^{100} \, + \, C_2 \, x^{99} \, - \, ....)$$              ....(1)
    Now E = $$C_0 \, C_1 \, - \, C-1 \, C_2 \, + \, C_2 \, C_3 \, - \, ....$$
    = -[Coefficient of $$x^{100}$$ in the product of  R.H.S. of (1)]   
    or     = -[Coefficient of $$(x^2)^{50}$$ in the product of  R.H.S. of (1)]    ....(2)
    Now 
    $$(x^2 \, - \, 1)^{101} \, = \, (-1)^{101} \, (1 \, - \, x^2)^{101} \, = \, -(1 \, - \, x^2)^{101}$$
    $$\therefore \, \, \, $$Coefficient of $$(x^2)^{50}$$
    = - $$(- 1)^{50} \, \, ^{101}C_{50} \, = \, - ^{101}C_{50}$$     ...(3)
    $$\therefore \, \, \, \, E \, = \, (-) \, (-) \, ^{101}C_{50} \, = \, ^{101}C_{50} \, by \, (2) \, and \, (3)$$
  • Question 5
    1 / -0
    The coefficient of $${x^6}.{y^{ - 2}}$$ in the expansion of $${\left( {\dfrac{{{x^2}}}{y} - \dfrac{y}{x}} \right)^{12}}$$ is
    Solution
    solve:

    Given, $$\left(\frac{x^{2}}{y}-\frac{y}{x}\right)^{12}$$ we have to find cofficient of $$x^{6} y^{-2}$$ in its expansion.

    It General term is

    $$T_{r+1}={ }^{12}{ }_{C_{r}}\left(\frac{x^{2}}{y}\right)^{12-r}\left(\frac{-y}{x}\right)^{r}$$

    $$\Rightarrow T_{r+1}={ }^{12} C_{r}(-1)^{r} \mathscr{x}^{(24-2 r-r)} \cdot y^{(r-12+r)}$$

    according to condition

     $$24-3 r=6 \\$$

    $$\Rightarrow  3 r=18 \\$$

    $$\Rightarrow  r=6$$

    and

     $$2 r-12=-2 \\$$

    $$\Rightarrow  2 r=10 \\$$

    $$\Rightarrow  r=5$$

    we find different Value of $$r$$ to get

    require power of $$x$$ and $$y$$

    Hence $$\left(x^{6} y^{-2}\right)$$ not exist

    So, it's cofficient is 0 .
  • Question 6
    1 / -0
    The co-efficient of $${x^{53}}$$ in the expression $$\sum\limits_{m = 0}^{100} {{}^{100}} {c_m}{(x - 3)^{100 - m}}{2^m}\,$$ is
    Solution
    $$S=\sum_{m=0}^{100}{^{100}C_{m}{\left(x-3\right)}^{100-m}{2}^{m}}$$

    $$^{100}C_{m}{\left(x-3\right)}^{100-m}{2}^{m}$$ is general term of $${\left(x-3+2\right)}^{100}$$

    Hence $$\sum_{m=0}^{100}{^{100}C_{m}{\left(x-3\right)}^{100-m}{2}^{m}}={\left(x-3+2\right)}^{100}$$

    $${\left(x-1\right)}^{100}={\left(1-x\right)}^{100}$$

    Hence coefficient of $${x}^{53}$$ in 
    $${\left(1-x\right)}^{100}={\left(-1\right)}^{53}\,^{100}C_{53}=-^{100}C_{53}$$
  • Question 7
    1 / -0
    If $${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\,\,$$ then the sum of the series $$\,1 + {}^n{C_r} + {}^n + 1{C_r} + {}^{n + 2}{C_3} + ......... + {}^{n + r - 1}{C_r}$$ is 
    Solution
    Given that:
    $$S=1+^nc_1+^{n+1}c_2+^{n+2}c_3+.....+^{n+r-1}c_r$$
    $$S=^nc_n+^nc_1+^{n+1}c_2+^{n+2}c_3+.....+^{n+r-1}c_r$$

    Now, using the property $$^nc_r=^nc_{n-r}$$

    $$S=^nc_n+^nc_{n-1}+^{n+1}c_{n-1}+^{n+2}c_{n-1}+.....+^{n+r-1}c_{n-1}$$

    $$S=^{n+1}c_n+^{n+1}c_{n-1}+^{n+2}c_{n-1}+.....+^{n+r-1}c_{n-1}$$

    $$S=^{n+2}c_n+^{n+2}c_{n-1}+.....+^{n+r-2}c_{n-1}+^{n+r-1}c_{n-1}$$
    $$...$$
    $$....$$

    $$S=^{n+r-1}c_{n}+^{n+r-1}c_{n-1}$$

    $$S=^{n+r}c_n$$

    Again using the property $$^nc_r=^nc_{n-r}$$, we get

    $$S=\ ^{n+r}c_r$$.
  • Question 8
    1 / -0
    In the expression of $$\left( {{2^x} + \frac{1}{{{4^x}}}} \right)^n\,$$ ratio  of 2nd and third terms is given by$$\,{t_3}/{t_2} = 7$$ and the sum of the co-efficients of 2nd and 3rd term is $$36,$$ then the value of $$x$$ is 
    Solution

    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.

  • Question 9
    1 / -0
    The coefficient $${x^n}$$ in the expression of $${\left( {1 + x} \right)^{2n}}$$ and $${\left( {1 + x} \right)^{2n - 1}}$$ are in the ratio.
    Solution
    We know that,
    General term of $$(a+b)^n$$ is
    $$T_r+1=^nC_r a^{n-r},b^r$$
    For $$(1+x)^{2n}$$
    General term of $$(1+x)^{2n}$$
    Put $$a=1,b=x,n=2n$$
    $$T_r+1=^{2n}C_r(1)^{2n-r}.(x)^r$$
    $$TY_r+1=^{2n}C_r.(x)^r$$......(i)
    For coedfficient of $$x^n$$
    Put $$r=n$$ in $$eq^n$$(i)
    $$T_r+1=^{2n}C_nx^n$$
    Coefficient of $$x^n=^{2n}C_n$$
    $$\therefore 2^nC_n=\dfrac{2n!}{n!(2n-n)!}$$
    $$=\dfrac{2n!}{n!(n)!}$$

    For $$(1+x)^{2n-1}$$
    General term of $$(1+x)^{2n-1}$$
    Put $$a=1,b=x,n=2n-1$$
    $$T_r+1=^{2r-1}C^r.(1)^{2n-1-r}.x^r$$
    $$T_r+1=^{2n-1}C_rx^r$$.....(ii)
    For coefficient of $$x^n$$
    Put $$r=n$$ in equation (ii)
    $$T_r+1=^{2n-1}C_n\,x^n$$
    Coefficient of $$x^n$$
    $$=2^{n-1}C_n\times 2$$
    $$=2\times \dfrac{(2n-1)!}{n!(2n-1-n)!}$$
    $$=2\times \dfrac{(2n-1)!}{n!(n-1)!}$$
    Multiply and divide by $$n$$, 
    we get,
    $$=\dfrac{2n(2n-1)!}{n!n(n-1)!}$$
    $$=\dfrac{(2n)!}{n!n!}$$
    Hence, ration is $$1:2$$
  • Question 10
    1 / -0
    The coefficient of $$x^{8}$$ in the polynomial $$\left( x-1 \right) \left( x-2 \right) \left( x-3 \right) ...\left( x-10 \right) $$ is
    Solution
    $$(x-1)(x-2)\Rightarrow x^2-3x+2 \Rightarrow x^2 -(2+1)x+(2+1)$$
    $$(x-1)(x-2)(x-3)\Rightarrow x^3-(1+2+3)x^2 +(1\times 2+1\times 3+2\times 3)x +(1\times 2\times 3)$$
    $$(x-a_1)(x-a_2)....(x-a_n)=x^n-(a_1+a_2+...a_n)x^{n-1} +(a_1 a_2....a_n)$$
    $$(x-1)(x-2)(x-3)....(x-10)$$
    $$x^8 \Rightarrow [2+3+4+...+10]+2 [3+4+....+10]+...9\times 10$$
    $$=1\times 54+2\times 52+3\times 49+....9\times 10$$
    $$=\boxed {1320}\Rightarrow $$ Required coefficient of $$x^{8}$$
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