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Binomial Theorem & its Simple Applications Test - 3

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Binomial Theorem & its Simple Applications Test - 3
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  • Question 1
    2 / -0.83

    The coefficient of second, third and fourth terms in the binomial expansion of  (1+x)n (‘n ’, a + ve integer) are in A.P.., if n is equal to

    Solution

  • Question 2
    2 / -0.83

    The coefficient of  x3 in the binomial expansion of   

    Solution

    T(r+1) = 11 Cr x(11-2r) (m/r)r (-1)r
    = 11 Cr x(11-2r) mr (-1)r
    Coefficient of x3 = 11 - 2r = 3
    8 = 2r  
    r = 4
    T5 = 11 C4 x3 m4 (-1)
    Coefficient of x3 = 11 C4 m4
    = 330 m4

  • Question 3
    2 / -0.83

    The coefficient of  x17 in the expansion of (x- 1) (x- 2) …..(x –18) is

    Solution

    The coefficient of x17 is given by  
    −1 + (−2) + (−3) + ….. (−18)
    = −1 −2 −3 ….. −18
    = −(18(18+1))/2
    = −9(19)
    = −171

  • Question 4
    2 / -0.83

    If the coefficients of (r +1)th term and (r + 3)th term in the expansion of  (1+x)2n be equal then

    Solution

    Tr+1 = 2n Cr(x)r (1)2n-r
    Tr+3 = 2n Cr+2 (x)r+2 (1)2n-r-2
    Tr+1 : Tr+3  
    = 2n Cr = 2n Cr+2
    =>2n!/(2n-r)!r! = 2n!/(r+2)!(2n-r-2)!
    =>1/(2n-r)(2n-r-1) = 1/(r+2)(r+1)
    =>(r+2)(r+1) = (2n-r)(2n-r-1)
    =>r2 + 3r + 2 = 4n2 - 2nr - 2n - 2nr + r^2 + r
    =>0 = 4n2 - 4nr - 2n + r - 3r - 2
    =>0 = 4n2 - 4nr - 2n + r - 3r - 2 - 2 + 2
    =>0 = 4n2 - 4nr - 4 - [2n + 2r - 2]
    =>4n(n-r-1) -2(n-r-1)
    Therefore n - r - 1 = 0
    =>n = r+1

  • Question 5
    2 / -0.83

    If  x = 9950 +1005  and  y = (101)50 , then

  • Question 6
    2 / -0.83

    The greatest coefficient in the expansion of  (1+x)12  is

  • Question 7
    2 / -0.83

    In the expansion of  (1+x)60 , the sum of coefficients of odd powers of x is

    Solution

     (1+x)n = nC0+ nC1x+ nC2x2 +...+ nC0xn ....(1)
    (1 −x)n = nC0 −nC1x+ nC2x2 −...+ nC0 xn ....(2)
    ⇒(1+x)n −(1 −x)n = nC0+ nC1x+ nC2x2 +...+ nC0xn −nC0+ nC1x −nC2x2 +...+ nC0xn
     =2(nC1x+nC3x3 +....+xn )

    Given: n=60 and put x=1
    ⇒(1+1)60 −(1 −1)60
     =260
    ⇒260 =2(nC1x+nC3x3 +....+xn )

    ∴Sum of odd powers of x in the expansion is = nC1x+nC3x3 +....+xn =260 −1 =259

  • Question 8
    2 / -0.83

    Let n  ∈ Q an n  ∉N,n ≠0, a  > 0, then the expansion of  (a+x)n in powers of x is valid if

  • Question 9
    2 / -0.83

    The expansion of  , in powers of x, is valid if

    Solution

    In case of negative or fractional power, expansion (1+x)^n is valid only when |x| <1
    (6 - 3x)-1/2
    = (6-1/2 (1 - x/2)-1/2 )
    So, this equation exists only when |x/2| <1
    |x| <2

  • Question 10
    2 / -0.83

    The middle term in the expansion of  (1+x)2n is

    Solution

    Middle term in the expansion of (1+x)2n ; is  
    =tn+1 = 2nCn.1(2n −n) .xn
    = {(2n)!.xn }/(2n −n)!n!
    = {{2n(2n −1)(2n −2)(2n −3)....4 ×3 ×2 ×1}/n! n!}/xn
    = {{2n [n(n −1)(n −2)....×2 ×1][(2n −1)(2n −3)....3 ×1]}/n! n!}xn
    = [[(2n −1)(2n −3)....3 ×1]/n!] 2n xn

  • Question 11
    2 / -0.83

    The coefficient of x99  in (x+1)(x+3)(x+5)………..(x+199) is

    Solution

    (x+1)(x+3)(x+5)(x+7)...(x+199)
    We have (x+1)(x+3)=x2 +(1+3)x+3
    (x+1)(x+3)(x+5)=x^3+(1+3+5)x2 +(3+15+5)x+15
    here Maximum power of x can be 100
    For x^99 we have to multiply x from 99 terms and constant from remaning 1 term
    So, when constant will be taken from (x+1) coff. of $$x99  will be 1
    when constant term will be taken from (x+3) coff. of x99 will be 3
    Similarly, When constant term will be taken from (x+199) coff. of x99 will be 199.
    ∴coefficient of x^99will be sum of all these coffecient=1+3+5+..+199

  • Question 12
    2 / -0.83

    If the expansion of in powers of x contains the term  x2r , then n −2r is

    Solution

    T(r+1) =  n Cr xn-r ar
    T(r+1) =  n-5 Cr xn-5-r (1/x4 )r
    = n-5 Cr xn-5-r (1/x4r )
    = n-5 Cr xn-5-5r  
    =>n - 5 - 5r = 2r
    =>n - 2r = 5(r + 1)

  • Question 13
    2 / -0.83

    Solution

  • Question 14
    2 / -0.83

    If rth ,(r+1)th and (r+2)th terms in the expansion of  (1+x)n are in A.P. then

  • Question 15
    2 / -0.83

    Coefficient of  a2 b5 in the expansion of  (a+b)3 (a −2b)4 is

    Solution

  • Question 16
    2 / -0.83

    In the expansion of(1+x)11 , the  5th term is 24 times the  3rd term . The value of x is

  • Question 17
    2 / -0.83

    The sum of coefficients in the expansion of  (x+2y+z)n  is (n being a positive integer)

  • Question 18
    2 / -0.83

    If the rth term in the expansion of   contains  x4  then r is equal to

    Solution

  • Question 19
    2 / -0.83

    If a + b = 1, then   is equal to  

  • Question 20
    2 / -0.83

    The 1st three terms in the expansion of (4 + x)3/2  are

  • Question 21
    2 / -0.83

    The coefficients of  xn in the expansion of  (1+2x + 3x2 + ........)1/2 is  

  • Question 22
    2 / -0.83

    If z =   + ........, then z2 + 2z is equal to

  • Question 23
    2 / -0.83

    The index of the power of x that occurs in the  7th term from the end in the expansion of  

  • Question 24
    2 / -0.83

    The index of the power of x that occurs in the  6th term in the expansion of  

    Solution

  • Question 25
    2 / -0.83

    If in the expansion of(1+x)43 , the coefficients of (2r+1)th and (r+2)th  terms are equal, then r is equal to

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