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Sequences and Series Test - 48

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Sequences and Series Test - 48
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  • Question 1
    1 / -0
    $$496 : 204 : : 329 : ?$$
  • Question 2
    1 / -0
    If $$x_{1},x_{2},....$$ are in H.P and $$x_{1},2,x_{20}$$ are in G.P, then $$\displaystyle \sum _{ 1 }^{ 19 }{ x_{r} .x_{r+1}}=$$ 
    Solution

  • Question 3
    1 / -0
    If $$S_n=\dfrac{7}{4.1.2}+\dfrac{10}{4^2.2.3}+\dfrac{13}{4^3.3.4}+....$$ then $$S_{\propto}$$ is equal to?
    Solution

  • Question 4
    1 / -0
    If $${ sin }^{ 4 }\alpha .{ cos }^{ 2 }\alpha =\sum _{ r=0 }^{ 3 }{ { C }_{ r } } .cos(2r\alpha )$$ then $${ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 }+{ C }_{ 3 }=$$
    Solution

  • Question 5
    1 / -0
    The value of  $$2 ^ { n } \{ 1.3 .5 \ldots \ldots ( 2 n - 3 ) ( 2 n - 1 ) \}$$  is
    Solution
    we have
    $$\begin{array}{l} { 2^{ n } }\left[ { 1.3.5........\left( { 2n-3 } \right) \left( { 2n-1 } \right)  } \right]  \\ Multiply\, and\, divide\, by\, 2.4.6....2n\, we\, get \\ \left[ { 1.3.5........\left( { 2n-3 } \right) \left( { 2n-1 } \right)  } \right] { 2^{ n } }=\left[ { 1.3.5........\left( { 2n-3 } \right) \left( { 2n-1 } \right)  } \right] { 2^{ n } }\times \frac { { 2.4.6....2n } }{ { 2.4.6....2n } }  \\ =\dfrac { { \left[ { 1.2.3.4.5.......\left( { 2n-1 } \right) \left( { 2n } \right)  } \right] { 2^{ n } } } }{ { 2.4.6......2n } }  \\ On\, \, \, rearraning\, the\, \, numerator,\, we\, get \\ =\dfrac { { \left[ { 1.2.3.4.5.......\left( { 2n-1 } \right) \left( { 2n } \right)  } \right] { 2^{ n } } } }{ { 2.4.6......2n } }  \\ =\dfrac { { \left[ { \left( { 2n } \right) ! } \right] { 2^{ n } } } }{ { 2.4.6......2n } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left\{ { \left( { 2n } \right) !=\left( { 2n } \right) \left( { 2n-1 } \right) ..........5.4.3.2.1 } \right\}  \\ Taking\, out\, 2\, common\, from\, deno\min  ator \\ =\dfrac { { \left[ { \left( { 2! } \right)  } \right] { 2^{ n } } } }{ { \left[ { 2.2.....n\, \, times } \right] \times \left[ { 1.2.3.....n\, \, times } \right]  } }  \\ =\dfrac { { \left( { 2n } \right) !{ 2^{ n } } } }{ { \left[ { { 2^{ n } } } \right] \times \left[ { \left( n \right) \left( { n+1 } \right) ......3.2.1 } \right]  } }  \\ =\dfrac { { \left( { 2n } \right) ! } }{ { n\left( { n-1 } \right) ......3.2.1 } }  \\ =\dfrac { { \left( { 2n } \right) ! } }{ { n! } }  \\ Hence,\, the\, option\, A\, is\, the\, \, correct\, \, answer. \end{array}$$
  • Question 6
    1 / -0
    The value of the series $$\dfrac{1}{3!}+\dfrac{2}{5!}+\dfrac{3}{7!}+..... $$  to $$\infty $$   is equal to 
    Solution
    $$\dfrac{0}{1!}+\dfrac{1}{3!}+\dfrac{2}{5!}+\dfrac{3}{7!}+\dfrac{4}{9!}+...=S$$

    $$S=\sum_{k=0}^{\infty}{\dfrac{k}{\left(2k+1\right)!}}$$

    $$=\dfrac{1}{2}\sum_{k=0}^{\infty}{\dfrac{2k}{\left(2k+1\right)!}}$$

    $$=\dfrac{1}{2}\sum_{k=0}^{\infty}{\dfrac{2k+1-1}{\left(2k+1\right)!}}$$

    $$=\dfrac{1}{2}\left[\sum_{k=0}^{\infty}{\dfrac{2k+1}{\left(2k+1\right)!}}-\sum_{k=0}^{\infty}{\dfrac{1}{\left(2k+1\right)!}}\right]$$

    $$=\dfrac{1}{2}\left[\sum_{k=0}^{\infty}{\dfrac{1}{\left(2k\right)!}}-\sum_{k=0}^{\infty}{\dfrac{1}{\left(2k+1\right)!}}\right]$$

    $$=\dfrac{1}{2}\sum_{k=0}^{\infty}{\dfrac{1}{\left(2k\right)!}}-\dfrac{1}{2}\sum_{k=0}^{\infty}{\dfrac{1}{\left(2k+1\right)!}}$$

    Now,$$e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+...$$

    $${e}^{-1}=1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+...$$

    $$\Rightarrow\,e+{e}^{-1}=2\left[1+\dfrac{1}{2!}+\dfrac{1}{4!}+...\right]=2\sum_{k=1}^{\infty}{\dfrac{1}{\left(2k\right)!}}$$

    $$\Rightarrow\,\sum_{k=1}^{\infty}{\dfrac{1}{\left(2k\right)!}}=\dfrac{1}{2}\left(e+{e}^{-1}\right)$$

    $$e-{e}^{-1}=2\left[1+\dfrac{1}{3!}+\dfrac{1}{5!}+...\right]=2\sum_{k=1}^{\infty}{\dfrac{1}{\left(2k+1\right)!}}$$

    $$\Rightarrow\,\sum_{k=1}^{\infty}{\dfrac{1}{\left(2k+1\right)!}}=\dfrac{1}{2}\left(e-{e}^{-1}\right)$$

    $$\therefore\,S=\dfrac{1}{2}\left(\dfrac{e+{e}^{-1}}{2}\right)-\dfrac{1}{2}\left(\dfrac{e-{e}^{-1}}{2}\right)$$

    $$=\dfrac{1}{4}\left(e+{e}^{-1}-e+{e}^{-1}\right)$$

    $$=\dfrac{1}{4}\left(2{e}^{-1}\right)$$

    $$=\dfrac{1}{2e}$$
  • Question 7
    1 / -0
    The sum of the series
      $$\dfrac { 9 }{ 5^{ { 2 } }\cdot 2.1 } +\dfrac { 13 }{ 5^{ { 3 } }\cdot 3.2 } +\dfrac { 17 }{ 5^{ { 4 } }\cdot 4.3 } ..........$$  Infinite terms
    Solution
    $$\frac{9}{5^{2} \cdot 2 \cdot 1}+\frac{13}{5^{3} \cdot 3 \cdot 2}+\frac{17}{5^{4}  \cdot 4 \cdot 3}................ \quad \infty \text { term }$$

    The general term of the above series,
    for, $$r \geqslant 2$$
    General term of the numerator

    $$T_{r}^{\prime}=4 r+1$$

    General term of the denominater
    $$T_{r^{\prime \prime}}=\frac{1}{r(r-1)}$$

    Then
    $$T_{r}=\frac{4 r+1}{r(r-1) \cdot 5^{r}} \quad$$ where, $$r \geqslant 2$$
    $$=\frac{5 r-(r-1)}{5^{r}r(r-1)}$$
    $$=\frac{5 r}{5^{r} \cdot(r(r-1)}-\frac{r-1}{r \cdot(r-1)}$$

    $$\Rightarrow \frac{1}{5^{r-1}(r-1)}-\frac{1}{5^{r} \cdot r}$$

    Now, putting the $$r \geqslant 2$$ in above equation

    $$T_{2}=\frac{1}{5 \cdot 1}-\frac{1}{5^{2} \cdot 2}$$

    $$T_{3}=\frac{1}{5^{2} \cdot 2}-\frac{1}{5^{3} \cdot 3}$$
    $$\because \text { every last term is gettimg cancelled }$$

    $$\therefore \text { o } \quad \sum_{r=2}^{\infty} t_{r}=\frac{1}{5}$$
    Option B
  • Question 8
    1 / -0
    The sum to infinity of the series $$1+\dfrac {2}{3}+\dfrac { 6 }{ { 3 }^{ 2 } } +\dfrac { 10 }{ { 3 }^{ 3 } } +\dfrac { 14 }{ { 3 }^{ 4 } }+.... $$ is
    Solution
    Let $$S = 1 + \cfrac{2}{3} + \cfrac{6}{{3}^{2}} + ..... \quad \quad ..... \left( 1 \right)$$
    $$\cfrac{S}{3} = \cfrac{1}{3} + \cfrac{2}{{3}^{2}} + \cfrac{6}{{3}^{3}} + ..... \quad \quad ..... \left( 2 \right)$$

    Subtracting equation $$\left( 2 \right)$$ from $$\left( 1 \right)$$, we have
    $$S - \cfrac{S}{3} = \left( 1 + \cfrac{2}{3} + \cfrac{6}{{3}^{2}} + ..... \right) - \left( \cfrac{1}{3} + \cfrac{2}{{3}^{2}} + \cfrac{6}{{3}^{3}} + ..... \right)$$
    $$\cfrac{2S}{3} = 1 + \cfrac{1}{3} + \cfrac{4}{{3}^{2}} + \cfrac{4}{{3}^{3}} + \cfrac{4}{{3}^{3}} + .....$$
    $$S = \cfrac{3}{2} + \cfrac{1}{2} + \cfrac{2}{3} + \cfrac{2}{{3}^{2}} + \cfrac{2}{{3}^{3}} + .....$$
    $$S = 2 + \cfrac{2}{3} + \cfrac{2}{{3}^{2}} + \cfrac{2}{{3}^{3}} + .....$$

    Since the above series is in G.P. of infinite terms with common ration $$\cfrac{1}{3}$$.
    $$\therefore S = \cfrac{2}{1 - \cfrac{1}{3}} \\ \Rightarrow S = \cfrac{2 \times 3}{2} = 3$$

    Hence, this is the answer.
  • Question 9
    1 / -0
    The sum to  $$50$$  terms of the series  $$\dfrac { 3 } { 1 ^ { 2 } } + \dfrac { 5 } { 1 ^ { 2 } + 2 ^ { 2 } } + \dfrac { 7 } { 1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } } + \ldots$$  is
    Solution
    $$\begin{array}{l} \frac { 3 }{ { { 1^{ 2 } } } } +\frac { 5 }{ { { 1^{ 2 } }+{ 2^{ 2 } } } } +\frac { 7 }{ { { 1^{ 2 } }+{ 2^{ 2 } }+{ 3^{ 2 } } } } + \\ { T_{ r } }=\frac { { 2r+1 } }{ { { 1^{ 2 } }+{ 2^{ 2 } }+.......{ r^{ 2 } } } }  \\ =\frac { { \left( { 2r+1 } \right) \times 2 } }{ { r\left( { r+1 } \right) \left( { 2r+1 } \right)  } }  \\ =\frac { 2 }{ { r\left( { r+1 } \right)  } }  \\ \sum  _{ r=1 }^{ 50 }{ { T_{ r } } }=2\sum  _{ r=1 }^{ 50 }{ \frac { { \left( { r+1 } \right) -r } }{ { r\left( { r+1 } \right)  } }  }=2\sum  _{ r=1 }^{ 50 }{ \left[ { \frac { 1 }{ r } -\frac { 1 }{ { r+1 } }  } \right]  } \\ =2\left[ { \left( { 1-\frac { 1 }{ 2 }  } \right) +\left( { \frac { 1 }{ 2 } -\frac { 1 }{ 3 }  } \right) +\left( { \frac { 1 }{ 3 } -\frac { 1 }{ 4 }  } \right) +........+\left( { \frac { 1 }{ { 50 } } -\frac { 1 }{ { 51 } }  } \right)  } \right]  \\ =2\left[ { 1-\frac { 1 }{ { 51 } }  } \right]  \\ we\, \, get, \\ =\frac { { 100 } }{ { 51 } }  \\ Hence, \\ option\, \, A\, \, is\, correct\, \, answer. \end{array}$$
  • Question 10
    1 / -0
    If $$an =\displaystyle \sum_{r =0}^{n} \dfrac{1}{^{n}C_{r}}$$ then $$\displaystyle \sum_{r=0}^{n} \dfrac{r}{^{n}C_{r}}$$ equals
    Solution
    $$ \begin{aligned} \text { Solution- Given that } \\ \qquad a_{n} &=\sum_{r=0}^{n} \frac{1}{n_{n}} \\ S &=\sum_{r=0}^{n} \frac{n}{{ }^{n} C_{\mu}} \\ &=\sum_{n=0}^{n} \frac{n-(n-\mu)}{{ }^{n} C_{n}}=\sum_{n=0}^{n} \frac{n}{{ }^{n} C_{\mu}}-\sum_{n=0}^{n} \frac{n-n}{{ }^{n} C_{\mu}} \end{aligned} $$


    As we know that


    $${ }^{n} C_{n}={ }^{n} C_{n-\mu}$$

    $$ \Rightarrow S=n \sum_{r=0}^{n} \frac{1}{n_{n}}-\sum_{n=0}^{n} \frac{n-n}{n_{n-n}} $$
    $$ \begin{array}{l} S=n \cdot(a n)-S \\ 2 S=n(a n) \\ S=\frac{1}{2} n a_{n} \\ \text { Hence, }(C) \text { is the correct option. } \end{array} $$

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