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Sequences and Series Test - 62

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Sequences and Series Test - 62
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  • Question 1
    1 / -0
    40280625,732375,16275,465,18.6,1.24,?40280625, 732375, 16275, 465, 18.6, 1.24,?
    Solution
    40280625,732375,16275,465,18.6,1.2440280625, 732375,16275,465,18.6,1.24
    40280625732375=55\cfrac{40280625}{732375}=55
    73237516275=45\cfrac{732375}{16275}=45
    16275465=35\cfrac{16275}{465}=35
    46518.6=25\cfrac{465}{18.6}=25
    18.61.24=15\cfrac{18.6}{1.24}=15
    1.24x=5\cfrac{1.24}{x}=5
    x=0.248x=0.248

  • Question 2
    1 / -0
    435,354,282,219,165,?435, 354, 282, 219, 165, ?
    Solution
    16545=x\therefore 165-45=x
    x=120x=120.

  • Question 3
    1 / -0
    For a positive integer n, let fn(θ)=(2cosθ+1)(2cosθ1)(2cos2θ1)(2cos(22)θ1)......(2cos(2n1)θ1)f_n(\theta)=(2\cos \theta+1)(2\cos \theta -1)(2\cos 2\theta -1)(2\cos (2^2)\theta -1)......(2\cos (2^{n-1})\theta -1). Which one of the following hold(s) not good?
    Solution
    f(θ )=(2cos θ+1)(2cos θ1)(2cos 2θ1)(2cos (22)θ1)......(2cos (2n1)θ1).=(2cos θ+1)+(2cos θ1)....(2cos2n1θ+1) Now,fn(θ )=2cos2nθ+1f2(π 6 )=2cos (4×π 6 )+1=2cos 2π 3+1=0f3(π 8 )=2cos (8×π 8 )+1=1f4(π 32 )=2cos (16×π 32 )+1=1f5(π 128 )=2cos (32×π 128 )+1=1+2 Hence,theoptionDisthecorrectanswer.\begin{array}{l} f\left( \theta  \right) =\left( { 2\cos  \theta +1 } \right) \left( { 2\cos  \theta -1 } \right) \left( { 2\cos  2\theta -1 } \right) \left( { 2\cos  \left( { { 2^{ 2 } } } \right) \theta -1 } \right) ......\left( { 2\cos  \left( { { 2^{ n-1 } } } \right) \theta -1 } \right) . \\ =\left( { 2\cos  \theta +1 } \right) +\left( { 2\cos  \theta -1 } \right) ....\left( { 2\cos { 2^{ n-1 } } \theta +1 } \right)  \\ Now, \\ { f_{ n } }\left( \theta  \right) =2\cos { 2^{ n } } \theta +1 \\ { f_{ 2 } }\left( { \frac { \pi  }{ 6 }  } \right) =2\cos  \left( { 4\times \frac { \pi  }{ 6 }  } \right) +1 \\ =2\cos  \frac { { 2\pi  } }{ 3 } +1=0 \\ { f_{ 3 } }\left( { \frac { \pi  }{ 8 }  } \right) =2\cos  \left( { 8\times \frac { \pi  }{ 8 }  } \right) +1=-1 \\ { f_{ 4 } }\left( { \frac { \pi  }{ { 32 } }  } \right) =2\cos  \left( { 16\times \frac { \pi  }{ { 32 } }  } \right) +1=1 \\ { f_{ 5 } }\left( { \frac { \pi  }{ { 128 } }  } \right) =2\cos  \left( { 32\times \frac { \pi  }{ { 128 } }  } \right) +1=1+\sqrt { 2 }  \\ Hence,\, the\, option\, D\, is\, the\, correct\, answer. \end{array}
  • Question 4
    1 / -0
    If x=1+112+122+1+122+132+1+120192+120202x=\sqrt 1+\dfrac{1}{{1}^{2}}+\dfrac{1}{{2}^{2}}+\sqrt 1+\dfrac{1}{{2}^{2}}+\dfrac{1}{{3}^{2}}+\sqrt 1+\dfrac{1}{{2019}^{2}}+\dfrac{1}{{2020}^{2}} then 4040(2020x)4040(2020-x)
    Solution

  • Question 5
    1 / -0
    If ak=1K(K+1)a_{k}=\dfrac{1}{K(K+1)} for K=1,2,3,.....n,K=1, 2, 3,.....n, then (k=1nak)2=\left(\sum_{k=1}^{n}{a_{k}}\right)^{2}=
    Solution
    ak=1k(k+1) =1k1k+1 ( k=1nax)2( k=1n(1k1k+1 ) )2=(112 )+(1213 )     (1n1n+1 ) =(11n+1 )2=(n+11n+1 )2=n2(n+1) 2 \begin{array}{l} { a_{ k } }=\dfrac { 1 }{ { k\left( { k+1 } \right)  } } =\dfrac { 1 }{ k } -\dfrac { 1 }{ { k+1 } }  \\ { \left( { \sum  _{ k=1 }^{ n }{ { a_{ x } } } } \right) ^{ 2 } } \\ { \left( { \sum  _{ k=1 }^{ n }{ \left( { \dfrac { 1 }{ k } -\dfrac { 1 }{ { k+1 } }  } \right)  } } \right) ^{ 2 } } \\ =\left( { 1-\dfrac { 1 }{ 2 }  } \right) +\left( { \dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 }  } \right) \, \, \, \, \, \left( { \dfrac { 1 }{ n } \dfrac { { -1 } }{ { n+1 } }  } \right)  \\ ={ \left( { 1-\dfrac { 1 }{ { n+1 } }  } \right) ^{ 2 } } \\ ={ \left( { \dfrac { { n+1-1 } }{ { n+1 } }  } \right) ^{ 2 } } \\ =\dfrac { { { n^{ 2 } } } }{ { { { \left( { n+1 } \right)  }^{ 2 } } } }  \end{array}

    Hence, this is the answer.
  • Question 6
    1 / -0
    If x1,x2,x3,......,xn{ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },......,{ x }_{ n }\quad are the roots of xn+ax+b=0\quad { x }^{ n }+ax+b=0. then the value of (x1x2)(x1x3)(x1x4)......(x1xn)\quad \left( { x }_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) \left( { x }_{ 1 }-{ x }_{ 4 } \right) ......\left( { x }_{ 1 }-{ x }_{ n } \right)
    Solution

  • Question 7
    1 / -0
    C1C0+2C2C1+3C3C2+......+nCnCn1\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +2\dfrac { { C }_{ 2 } }{ { C }_{ 1 } } +3\frac { { C }_{ 3 } }{ { C }_{ 2 } } +......+\dfrac { n{ C }_{ n } }{ { C }_{ n-1 } } equals 
    Solution

  • Question 8
    1 / -0
    The value of 2+12+12+.....2+\dfrac{1}{2+\dfrac{1}{2+.....\infty}} is :
    Solution
    2+12+.... x=2+1x x21x=0x22x1=0x=+2±4+4 2=2±22 2 =1±2 \begin{array}{l} 2+\dfrac { 1 }{ { 2+.... } }  \\ x=2+\dfrac { 1 }{ x }  \\ x-2-\dfrac { 1 }{ x } =0 \\ { x^{ 2 } }-2x-1=0 \\ \dfrac { { x=+2\pm \sqrt { 4+4 }  } }{ 2 } =\dfrac { { 2\pm 2\sqrt { 2 }  } }{ 2 }  \\ =1\pm \sqrt { 2 }  \end{array}
  • Question 9
    1 / -0
    If (1+x+x2)=a0+a1x+a2x2+........a2nx2n\left( 1+x+{ x }^{ 2 } \right) ={ a }_{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+........{ a }_{ 2n }{ x }^{ 2n }, then the value of a0+a3+a6+.....{ a }_{ 0 }+{ a }_{ 3 }+{ a }_{ 6 }+..... is 
  • Question 10
    1 / -0
    The sum of the infinite series cot1(74)+cot1(194)+cot1(394)+cot1(674)+........\cot { ^{ 1 } } \left( \dfrac { 7 }{ 4 } \right) +\cot { ^{ -1 }\left( \dfrac { 19 }{ 4 } \right) } +\cot { ^{ -1 }\left( \dfrac { 39 }{ 4 } \right) } +\cot { ^{ -1 }\left( \dfrac { 67 }{ 4 } \right) } +........\infty is:
    Solution

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