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Binomial Theorem Test - 24

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Binomial Theorem Test - 24
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  • Question 1
    1 / -0
    In the expansion of $$\left (x+ \sqrt{x^{2}-1}\right )^{6}$$+ $$\left (x- \sqrt{x^{2}-1}\right )^{6}$$,the number of terms is
    Solution
    Let $$a=x,b=\sqrt{x^2-1}$$
    Now,
    $$(x+\sqrt{x^2-1})^6=(a+b)^6$$
    $$=^6C_0a^6+^6C_1a^5b+^6C_2a^4b^2+^6C_3a^3b^3+^6C_4a^2b^4+^6C_5ab^5+^6C_6 b^6$$
    And $$(x-\sqrt{x^2-1})^6= (a-b)^6$$
    $$=^6C_0a^6-^6C_1a^5b+^6C_2a^4b^2-^6C_3a^3b^3+^6C_4a^2b^4-^6C_5ab^5+^6C_6 b^6$$
    Now adding above two equations,
    $$(a+b)^6+(a-b)^6=2(^6C_0a^6+^6C_2a^4b^2+^6C_4a^2b^4+^6C_6b^6)$$
    Hence number of terms in the required expansion is $$4$$
  • Question 2
    1 / -0
    If $$\left ( 1+x \right )\left ( 1+x^{2} \right )\left ( 1+x^{4} \right )...\left ( 1+x^{128} \right )=\sum_{r=0}^{n}x^{r}$$ then $$n$$ is
    Solution
    The problem is to find the greatest power of $$x$$ in the expansion
    The greatest power is obtained when all the powers of $$x$$ in all the brackets add up
    Thus, $$ n= 1+2+4+ .. +128 $$
    $$ = \frac{1(2^{8}-1)}{2-1} = 256-1 = 255 $$
    Hence, A is correct.
  • Question 3
    1 / -0
    The number of terms with integral coefficients in the expansion of$$\left ( 7^{1/3}+5^{1/2}.x \right )^{600}$$ is
    Solution
    The number of integral terms will be
    $$1+\dfrac{100}{L.C.M(2,3)}$$
    $$=1+\dfrac{600}{6}$$
    $$=1+100$$
    $$=101$$
  • Question 4
    1 / -0
    The sum of coefficients in the binomial expansion of $$\displaystyle \left ( \frac{1}{x}+2x \right )^{n}$$is equal to $$6561$$.The constant term in the expansion is
    Solution
    Substituting $$x=1$$ to get the sum of the binomial coefficients
    we get
    $$(3)^{n}=6561$$
    Taking $$log_{3}$$ on both the sides, we get
    $$n=8$$
    Therefore the above question reduces to
    $$(x^{-1}+2x)^{8}$$
    The general term will be
    $$T_{r+1}=\:^{8}C_{r}x^{2r-8}2^{r}$$
    The term independent of $$x$$ will be given by
    $$2r-8=0$$
    $$r=4$$
    Substituting we get
    $$\:^{8}C_{4}2^{4}$$
    $$=16(\:^{8}C_{4})$$
  • Question 5
    1 / -0
    The coefficient of $$x^{49}$$ in the product $$(x-1)(x-3)\dots(x-99)$$ is
    Solution

    $$(x-α_{ 1 })(x-α_{ 2 })(x-\alpha_{ 3 })…(x-α_{ n })=x^{ n }+A_{ 1 }{ x }^{ n-1 }+A_{ 2 }{ x }^{ n-2 }+⋯+A_{ n }$$

    Where , 

    $$A_{ p }=(-1)^{ p }\sum(α_{ 1 }α_{ 2 }…α_{ p })\>;$$$$(p$$ terms at a time$$)$$

    Given,  

    $$(x-1)(x-3)\dots(x-99)$$ contains $$50$$ terms,

    Coefficient of $${ x }^{ 49 }=A_{ 1 }=(-1)^{ 1 }\sum(\alpha)=-(1+3+⋯+99)=-50\times50=-2500$$

    $$($$Sum of first $$n$$ odd numbers $$={ n }^{ 2 } )$$

  • Question 6
    1 / -0
    If the $$4th$$ term in the expansion is of $$ \left (px+x^{-1} \right )^{m}$$ is $$2.5$$ for all $$x\epsilon R$$ then
    Solution
    The fourth term has a value 2.5 for all Real x.
    Hence
    $$T_{3+1}=\:^{m}C_{3}p^{m-3}x^{m-3}x^{-3}$$
    It has to be a term independent of x
    Therefore
    $$m-6=0$$
    $$m=6$$
    Subtituiting we get
    $$\:^{6}C_{3}p^{3}=2.5$$
    $$20p^{3}=2.5$$
    $$p=0.5$$
  • Question 7
    1 / -0
    The number of real negative terms in the binomial expansion of $$\left ( 1+ix \right )^{4n-2},$$ $$n\epsilon N,$$ $$x>0,$$ is
    Solution
    $$(1+ix)^{4n-2}$$

    $$=((1+ix)^{2})^{2n-1}$$

    $$=(1-x^{2}+2ix)^{2n-1}$$

    $$=[(1-x^{2})+i(2x)]^{2n-1}$$
    Total number of terms will be $$2n-1+1=2n$$.
    Hence the number of real negative terms will therefore be 
    $$=\dfrac{2n}{2}$$
    $$=n$$.
  • Question 8
    1 / -0
    The number of terms whose values depends on $$x$$ in the expansion of $$\displaystyle\left (  x^{2}-2+\frac{1}{x^{2}}\right )^{n}$$ is
    Solution
    $$(x^2-2+\frac{1}{x^2})^{n}$$
    $$=[(x-\frac{1}{x})^{2}]^{n}$$
    $$=(x-\frac{1}{x})^{2n}$$
    Hence there will be 2n+1 terms.
    The middle term i.e $$n+1 ^{th}$$ term will be independent of x.
    Hence total number of terms, dependent on x will be
    $$2n+1-(1)$$
    $$=2n$$ terms.
  • Question 9
    1 / -0
    The absolute value of middle term in the expansion of $$\displaystyle \left ( 1-\frac{1}{x} \right )^{n}.(1-x)^{n}$$ is
    Solution
    Given, $$(1-\dfrac{1}{x})^{n}(1-x)^{n}$$
    $$=\dfrac{1}{x^{n}}(x-1)^{n}(1-x)^{n}$$
    $$=\dfrac{(-1)^{n}}{x^{n}}[1-x]^{2n}$$
    Therefore the middle term will be
    $$T_{n+1}=\:^{2n}C_{n}x^{n}\left(\dfrac{(-1)^{n}}{x^{n}}\right)$$
    $$=(-1)^{n}\:^{2n}C_{n}$$
  • Question 10
    1 / -0
    The middle term in the expansion of $$\left ( \displaystyle \frac{2x}{3}-\frac{3}{2x^{2}} \right )^{2n}$$ is
    Solution
    Middle term will be $$nth$$ term.
    Hence
    $$T_{n+1}$$
    $$=(-1)^{n}\:^{2n}C_{n}\dfrac{2}{3}^{n}.\dfrac{3}{2}^{n}.x^{-2n+n}$$
    $$=(-1)^{n}\:^{2n}C_{n}x^{-n}$$
    $$=(-1)^{n}\dfrac{2n!}{n!^{2}}x^{-n}$$
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