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Binomial Theorem Test - 42

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Binomial Theorem Test - 42
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  • Question 1
    1 / -0
    The coefficient of $$x^5$$ in the expansion of $$\dfrac{1+x^2}{a+x},|x| < 1$$, is 
    Solution

    Let $$ S= \dfrac{1+x^2}{a+x},|x| < 1$$

    $$S= (1+x^{2})(a+x)^{-1}$$

    $$S= a^{-1}(1+x^{2})(1+\frac{x}{a})^{-1}$$

     

    $$(1+x)^{n}= 1+nx+\frac{n(n-1)}{2!}x^{2}+\frac{n(n-1)(n-2)}{3!}x^{3}...$$

    Put $$n=-1$$ and $$x=\dfrac{x}{a}$$:

    $$\displaystyle \left(1+\dfrac{x}{a}\right)^{-1}= 1-\left(\dfrac{x}{a}\right)+\left(\dfrac{x}{a}\right)^{2}-\left(\dfrac{x}{a}\right)^{3}...=\sum_{0}^{\infty }\left(\frac{-1}{a}\right)^{n}x^{n}$$ 

     

    Substituting in $$S$$:

    $$ S=a^{-1}(1+x^{2})(1-(\frac{x}{a})+(\frac{x}{a})^{2}-(\frac{x}{a})^{3}...)$$

    $$\therefore$$  $$\displaystyle  S= a^{-1}\left(1-\left(\frac{x}{a}\right)+\sum_{2}^{\infty }\left[\left(\frac{-1}{a}\right)^{n}+\left(\frac{-1}{a}\right)^{n-2}\right]x^{n}\right)$$


    The coefficient of $$x^{5}$$ (Say $$\beta $$) in $$S= a^{-1}\left[\left(\dfrac{-1}{a}\right)^{5}+\left(\dfrac{-1}{a}\right)^{3}\right] $$

                      $$\therefore \beta =(-1)\left[\left(\dfrac{1}{a}\right)^{6}+\left(\dfrac{1}{a}\right)^{4}\right]$$ 

  • Question 2
    1 / -0
    If the ratio of the seventh term from the beginning of the binomial expansion of $${ \left( { 2 }^{ 1/3 }+\cfrac { 1 }{ { 3 }^{ 1/3 } }  \right)  }^{ x }$$ to the seventh term from its end is $$\cfrac { 1 }{ 6 } $$, then the value $$x$$ is
    Solution
    We have
    $${ \left( { 2 }^{ 1/3 }+\cfrac { 1 }{ { 3 }^{ 1/3 } }  \right)  }^{ x }$$ 
    $$\because$$ 7th term of expansion from beginning
    $$={ _{  }^{ x }{ C } }_{ 6 }{ \left( { 2 }^{ 1/3 } \right)  }^{ x-6 }{ \left( \cfrac { 1 }{ { 3 }^{ 1/3 } }  \right)  }^{ 6 }$$
    And 7th term of expansion from end
    $$={ _{  }^{ x }{ C } }_{x- 6 }{ \left( { 2 }^{ 1/3 } \right)  }^{ 6 }{ \left( \cfrac { 1 }{ { 3 }^{ 1/3 } }  \right)  }^{ x-6 }$$
    Given :
    $$\quad \cfrac { 7th\quad term\quad expansion\quad from\quad beginning }{ 7th\quad term\quad expansion\quad from\quad end } =\cfrac { 1 }{ 6 } $$
    $$\Rightarrow \cfrac { { _{  }^{ x }{ C } }_{ 6 }{ \left( { 2 }^{ 1/3 } \right)  }^{ x-6 }{ \left( \cfrac { 1 }{ { 3 }^{ 1/3 } }  \right)  }^{ 6 } }{ { _{  }^{ x }{ C } }_{ x-6 }{ \left( { 2 }^{ 1/3 } \right)  }^{ 6 }{ \left( \cfrac { 1 }{ { 3 }^{ 1/3 } }  \right)  }^{ x-6 } } =\cfrac { 1 }{ 6 } $$
    $$\Rightarrow \cfrac { { _{  }^{ x }{ C } }_{ 6 }{ \left( { 2 }^{ 1/3 } \right)  }^{ x-12 } }{ { _{  }^{ x }{ C } }_{ 6 }{ \left( \cfrac { 1 }{ { 3 }^{ 1/3 } }  \right)  }^{ x-12 } } =\cfrac { 1 }{ 6 } $$
    $${ \left( { 6 }^{ 1/3 } \right)  }^{ x-12 }=\cfrac { 1 }{ 6 } $$
    $$\Rightarrow \quad { \left( \cfrac { 1 }{ 6 }  \right)  }^{ -1/3(x-12) }\quad =\cfrac { 1 }{ 6 } $$
    $$\therefore -\cfrac { 1 }{ 3 } (x-12)=1$$
    $$12-x=3$$
    $$\Rightarrow$$  $$x=9$$
  • Question 3
    1 / -0
    If $${ C }_{ 0 },{ C }_{ 1 },{ C }_{ 2 },.....{ C }_{ r }$$ are binomial coefficients in the expansion of $${(1+x)}^{n}$$ then
    $${ C }_{ 1 }-\cfrac { { C }_{ 2 } }{ 2 } +\cfrac { { C }_{ 3 } }{ 3 } -\cfrac { { C }_{ 4 } }{ 4 } +....{ \left( -1 \right)  }^{ n-1 }\cfrac { { C }_{ n } }{ n } $$ equals
    Solution
    Consider $${ (1-x) }^{ n }={ C }_{ 0 }-{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }-{ C }_{ 3 }{ x }^{ 3 }+......$$

    $$\Rightarrow 1-{ (1-x) }^{ n }={ C }_{ 1 }x-{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }-......$$

    integrating both side w.r.t $$x$$ with limit $$0$$ to $$1$$

    $$\displaystyle \therefore \int _{ 0 }^{ 1 }{ \left( { C }_{ 1 }x-{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }-..... \right) dx } =\int _{ 0 }^{ 1 }{ \cfrac { 1-{ (1-x) }^{ n } }{ 1-(1-x) }  } dx$$

    $$\displaystyle \Rightarrow \cfrac { { C }_{ 1 } }{ 1 } -\cfrac { { C }_{ 2 } }{ 2 } +\cfrac { { C }_{ 3 } }{ 3 } +....{ (-1) }^{ n }-1\cfrac { { C }_{ n } }{ n } =\int _{ 0 }^{ 1 }{ \cfrac { 1-{ x }^{ n } }{ 1-x }  } dx\quad $$

    $$\displaystyle \left( Using\quad \int _{ 0 }^{ a }{ f(x) } dx=\int _{ 0 }^{ a }{ f(a- } x)dx \right) $$

    $$\displaystyle =\int _{ 0 }^{ 1 }{ \left( 1+x+{ x }^{ 2 }+...+{ x }^{ n-1 } \right) dx } $$

    $$=1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } +...\cfrac { 1 }{ n } =\sum _{ r=1 }^{ n }{ \cfrac { 1 }{ r }  } $$
  • Question 4
    1 / -0
    The coefficient of x in the expansion of $$(1-ax)^{-1}(1-bx)^{-1}(1-cx)^{-1}$$ is?
    Solution
    $${ (1-ax) }^{ -1 }=1+{ ^{ -1 }C }_{ 1 }(-ax)+\dfrac { (-1)(-2) }{ 2! } { (-ax)^2}$$
    We can get coefficient of $$x$$ by taking$$x$$ from one bracket and $$ 1$$ from remaining.
    $$\therefore$$ Coefficient of $$x = (a).(1).(1)$$
                                     $$+(1).(b).(1)$$
                                     $$+(1).(1).(c)$$
    $$=a+b+c.$$
    Hence the answer is $$a+b+c.$$

  • Question 5
    1 / -0
    The sum of the co-efficients of all odd degree terms in the expansion of $$\left(x + \sqrt {x^{3} - 1}\right)^{5} + (x - \sqrt {x^{3} - 1})^{5}, (x > 1)$$ is
    Solution
    $$(x+\sqrt {x^3-1})^5+(x-\sqrt {x^3-1})^5$$

    $$(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5  -------(i)$$

    $$(a-b)^5=a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5  -------(ii)$$

    $$(a+b)^5+(a-b)^5=2[a^5+10a^3b^2+5ab^4]$$

                                  $$=2[x^5+10x^3(x^3-1)+5x(x^3-1)^4]$$

                                   $$=2[x^5+10x^6-10x^3+5x(x^6-2x^3+1)]$$

                                   $$=2x^5+20x^6-20x^3+10x^7-20x^4+10x$$

    Here all the co-efficient of the above equation are odd terms coefficients of even terms are cancelled out.
    So, sum of coefficients $$\Rightarrow2+20-20+10-20+10=2$$
  • Question 6
    1 / -0
    If $${C}_{r}$$ denotes the binomial coefficient $${ _{  }^{ n }{ C } }_{ r }$$ then $$\left( -1 \right) { C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+5{ C }_{ 2 }^{ 2 }+......(3n-1){ C }_{ n }^{ 2 }=\quad $$
    Solution
    The general term of above series is $$T_{r} = (3r-1)C_{r}^{2} $$

    $$(-1)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+(3n-1)C_{n}^{2} $$

    $$=\sum_{r=0}^{r=n}(3r-1)C_{r}^{2} $$

    $$=3\sum_{r=0}^{r=n}rC_{r}^{2} - \sum_{r=0}^{r=n}C_{r}^{2}$$

    $$=3S_{1} - S_{2}$$

    The binomial expansion of $$(1+x)^n$$ and $$(1+\dfrac{1}{x})^n$$ is given by

    $$(1+x)^n = C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n} = \sum_{r=0}^{r=n}C_{r}x^{r}$$                          ....[1]
    $$(1+\dfrac{1}{x})^n = C_{0}+C_{1}\dfrac{1}{x}+C_{2}\dfrac{1}{x^{2}}+....+C_{n}\dfrac{1}{x^{n}} = \sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}}$$                 ....[2]

    To find the value of $$S_{2}$$, we can multiply above two equations and then compare the coefficients of terms independent of x.
    Multiplying equation [1] and [2], we get
    $$(1+x)^n (1+\dfrac{1}{x})^{n} = \sum_{r=0}^{r=n}C_{r}x^{r}.\sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}}$$

    $$\implies \dfrac{(1+x)^{2n}}{x^{n}}  = \sum_{r=0}^{r=n}C_{r}x^{r}.\sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}} $$
    $$\implies \dfrac{(1+x)^{2n}}{x^{n}}=C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}$$ + terms containing x
    Therefore, comparing coefficients of $$x^{0}$$ in L.H.S and R.H.S, we get
     Coefficients of $$x^{0}$$ in R.H.S = Coefficients of $$x^{0}$$ in L.H.S
     $$\implies C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}=$$ coefficient of $$x^{0}$$ in $$\dfrac{(1+x)^{2n}}{x^{n}} $$
    $$\implies C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}= ^{2n}C_{n} = S_{2}$$                                               ....[3]

    $$(1+\dfrac{1}{x})^n = C_{0}+C_{1}\dfrac{1}{x}+C_{2}\dfrac{1}{x^{2}}+....+C_{n}\dfrac{1}{x^{n}}$$  
    Differentiating with respect to x, we get
    $$\implies n(1+\dfrac{1}{x})^{n-1}(\dfrac{-1}{{x^{2}}}) = 0+C_{1}\dfrac{-1}{x^{2}}+C_{2}\dfrac{-2}{x^{3}}+....+C_{n}\dfrac{-n}{x^{n+1}}$$  
    $$\implies n(1+\dfrac{1}{x})^{n-1}(\dfrac{1}{{x}}) = C_{1}\dfrac{1}{x}+C_{2}\dfrac{2}{x^{2}}+....+C_{n}\dfrac{n}{x^{n}}$$                            ....[4]

    To find the value of $$S_{1}$$, we can multiply equations [1] and [4], and then compare the coefficients of terms independent of x.
    Multiplying equation [1] and [4], we get
    $$n(1+\dfrac{1}{x})^{n-1}(\dfrac{1}{{x}})(1+x)^n=(C_{1}\dfrac{1}{x}+C_{2}\dfrac{2}{x^{2}}+....+C_{n}\dfrac{n}{x^{n}})( C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n} )$$  
    $$\implies n\dfrac{(x+1)^{2n-1}}{x^{n}}=(C_{1}\dfrac{1}{x}+C_{2}\dfrac{2}{x^{2}}+....+C_{n}\dfrac{n}{x^{n}})( C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n} )$$  
    Therefore, comparing coefficients of $$x^{0}$$ in L.H.S and R.H.S, we get
     Coefficients of $$x^{0}$$ in R.H.S = Coefficients of $$x^{0}$$ in L.H.S
     $$\implies C_{1}^{2}+2C_{2}^{2}+3C_{3}^{2}+...+nC_{n}^{2}=$$ coefficient of $$x^{0}$$ in $$n\dfrac{(x+1)^{2n-1}}{x^{n}} $$
    $$\implies C_{1}^{2}+2C_{2}^{2}+3C_{3}^{2}+...+nC_{n}^{2}= n(^{2n-1}C_{n}) = S_{1} $$
    Therefore,
    $$3S_{1} - S_{2} = 3n(^{2n-1}C_{n})-^{2n}C_{n}$$
    $$= 3n\dfrac{(^{2n}C_{n})}{2}-^{2n}C_{n}$$
    $$= (\dfrac{3n-2}{2})^{2n}C_{n}$$

    Hence, the answer is option (B)
  • Question 7
    1 / -0
    The value of $$\cfrac { { C }_{ 0 } }{ 1.3 } -\cfrac { { C }_{ 1 } }{ 2.3 } +\cfrac { { C }_{ 2 } }{ 3.3 } -\cfrac { { C }_{ 3 } }{ 4.3 } +.....{ \left( -1 \right)  }^{ n }\cfrac { { C }_{ n } }{ (n+1).3 } $$ is
    Solution

    Consider the expansion $${ \left( 1-x \right)  }^{ n }={ C }_{ 0 }-{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }-...{ \left( -1 \right)  }^{ n }{ x }^{ n }$$,

    Now, integrating the expansion with respect to $$x$$ from $$0$$ to $$1$$ gives,

    $$\dfrac { 1 }{ n+1 } =\dfrac { { C }_{ 0 } }{ 1 } -\dfrac { { C }_{ 1 } }{ 2 } +\dfrac { { C }_{ 2 } }{ 2 } -...{ \left( -1 \right)  }^{ n }\dfrac { { C }_{ n } }{ n+1 } \\ \Longrightarrow \dfrac { 1 }{ 3\left( n+1 \right)  } =\dfrac { { C }_{ 0 } }{ 1.3 } -\dfrac { { C }_{ 1 } }{ 2.3 } +\dfrac { { C }_{ 2 } }{ 3.3 } -...{ \left( -1 \right)  }^{ n }\dfrac { { C }_{ n } }{ \left( n+1 \right) .3 } $$

     

  • Question 8
    1 / -0
    Find the term of the expansion of $$\displaystyle\, \left ( \sqrt[3]{x^{-2}} + x \right )^7$$ containing $$x$$ in the second power.
    Solution
    Note:- The general term or $$(r+1)^{th}$$ term of $$(a+b)^{n}$$ is $${T}_{r+1}=^{ n }{ C }_{ r }{a}^{n-r}{b}^{r}$$
    $${T}_{r+1}=^{ 7 }{ C }_{ r }\left( \sqrt [ 3 ]{ { x }^{ -2 } }  \right) ^{ 7-r }\left(x \right)^{r}$$
    $$\Rightarrow {T}_{r+1}= ^{ 7 }{ C }_{ r }\left({x}^{-2}\right)^{{7-r}/{3}}{x}^{r}$$
    $$\Rightarrow {T}_{r+1}=^{ 7 }{ C }_{ r } {x}^{{-2}/{3}(7-r)+r}$$
    Now, if power of $$x$$ must be $$2$$, then $$r$$ is:-
    $$\therefore {x}^{{-2}/{3}(7-r)+r}= {x}^{2}$$
    $$\Rightarrow \cfrac{-2}{3}(7-r)+r=2$$
    $$\Rightarrow \cfrac{-14}{3}+\cfrac{2}{3}r+r=2$$
    $$\Rightarrow \cfrac{5r}{3}= \cfrac{20}{3}$$
    $$\Rightarrow r=4$$
    Therefore $${T}_{4+1}={T}_{5}$$ is the term which contains $${x}^{2}$$
  • Question 9
    1 / -0
    The numerically greatest terms in the expansion of $${ \left( 2x+5y \right)  }^{ 34 }$$ when $$x=3$$ & $$y=2$$ is
    Solution
    $$(2x+5y)^{34}$$ can be written as $$(2x)^{34}$$.$$(1+ \dfrac{5y}{2x})^{34}$$ 

    $$\rightarrow$$  or $$(2x)^{34}$$.$$(1+ \alpha)^{34}$$, where $$\alpha$$ = $$\dfrac{5y}{3x}$$

    Value of $$\alpha$$ at x=3, y=2 is equal to $$\dfrac{5.2}{2.3}=\dfrac{5}{3}$$
    So $$\alpha_{x=3,y=2}= \dfrac{5}{3}$$

    Now the numerically greatest term in the expansion of $$(2x+5y)^{34}$$ is the numerically greatest term in the expansion of $$(1+\alpha)^{34}$$
    We know that for any expansion $$(1+\alpha)^n$$, If the numerically greatest term is $$T_r$$, where r is an integer. 
    Then the value of r is $$\leq\dfrac {(n+1)|\alpha|}{|\alpha+1|}$$
    So to know the numerically greatest term in the given expansion at x=3, y=2 we need to find the integer value of
    r     $$\leq\dfrac {(n+1)|\alpha|}{|\alpha+1|}$$ at x=3, y=2, Here the value of n = 34

    So $$r\leq\dfrac{(34+1)|\dfrac{5}{3}|}{|\dfrac{5}{3}+1|}$$
    $$\rightarrow$$ $$\dfrac{35.5}{8} = \dfrac{175}{8} = 21.8$$

    As r is an integer so the closest value of integer less than 21.8 is 21.
    Hence the numerically greatest term will be 21th term or $$T_{21}$$
    So the correct option is A.
  • Question 10
    1 / -0
    If $$T_0, T_1, T_2, .....T_n$$ represent the term in the expansion of $$(x+a)^n$$, then the value of $$(T_0- T_2+ T_4- T_6+ .....)^2+(T_1-   T_3+ T_5+.....)^2$$ is
    Solution
    Given 
    $$T_{0},T_{1},T_{2},...............,T_{n}$$ represent terms in expansion of $$(x+a)^n$$

    $$(x+a)^n=T_{0}+T_{1}+T_{2}+T_{3}+.........+ T_{n}---------(1)$$
    When $$a=-a$$ we get
    $$(x-a)^n=T_{0}-T_{1}+T_{2}-T_{3}+.........+(-1)^n T_{n}------(2)$$

    Multiplying eq (1) and (2) and comparing with terms $$x^r$$ we get
    $$(x+a)^n\cdot (x-a)^n=(T_{0}-T_{2}+T_{4}-T_{6}+.........)^2+(T_{1}-T_{3}+T_{5}+.........)^2$$

    $$(x^2-a^2)^n=(T_{0}-T_{2}+T_{4}-T_{6}+.........)^2+(T_{1}-T_{3}+T_{5}+.........)^2$$ 

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