Sequences and Series Test

Result

Sequences and Series Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

The number of real roots of the equation $$1+a_{1}x+a_{2}x^{2}+....+a_{n}x^{n}=0$$ where $$|x| < \dfrac {1}{3}$$ and $$|a_{n}| < 2$$, is

A
$$n$$ if $$n$$ is even

B
$$0$$ for any natural number $$n$$

C
$$1$$ if $$n$$ is odd

D
$$None\ of\ these$$

Solution
Question 2
1 / -0

If $${\left( {20} \right)^{19}} + 2\left( {21} \right){\left( {20} \right)^{18}} + 3{\left( {21} \right)^2}{\left( {20} \right)^{17}} + ... + 20{\left( {21} \right)^{19}} = k{\left( {20} \right)^{19}}$$
then $$k$$ is equal to

A
400

B
100

C
441

D
420

Solution
Question 3
1 / -0

If $$(1+x)^{n}=C_{0}+C_{1}x+C_{2}x^{2}+....+...C_{n}x^{n}$$, then $$\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +\dfrac { { 2C }_{ 2 } }{ { C }_{ 1 } } +\dfrac { { 3C }_{ 3 } }{ { C }_{ 2 } } +....+\dfrac { { nC }_{ n } }{ { C }_{ n-1 } } =$$

A
$$\dfrac {n(n-1)}{2}$$

B
$$\dfrac {n(n+2)}{2}$$

C
$$\dfrac {n(n+1)}{2}$$

D
$$\dfrac {(n-1)(n-2)}{2}$$

Solution
Question 4
1 / -0

The sum of the series $$^{20}C_{0}-^{20}C_{1}+^{20}C_{2}-^{20}C_{3}+-..+^{20}C_{10}$$ is$$\dfrac{1}{2}^{20}C_{10}$$

A
$$0$$

B
$$-^{20}C_{10}$$

C
$$^{20}C_{10}$$

D
$$\dfrac{1}{2}^{20}C_{10}$$

Solution

The sum of the series
$$\begin{array}{l} ^{ 20 }{ C_{ 0 } }{ -^{ 20 } }{ C_{ 1 } }{ +^{ 20 } }{ C_{ 2 } }{ -^{ 20 } }{ C_{ 3 } }+{ .....^{ 20 } }{ C_{ 10 } } \\ { \left( { 1+x } \right) ^{ 20 } }{ =^{ 20 } }{ C_{ 0 } }{ +^{ 20 } }{ C_{ 1 } }x{ +^{ 20 } }{ C_{ 2 } }{ x^{ 2 } }+{ ....^{ 20 } }{ C_{ 10 } }{ x^{ 10 } } \\ putx=-1 \\ { \left( { 1-1 } \right) ^{ 20 } }{ =^{ 20 } }{ C_{ 0 } }{ +^{ 20 } }{ C_{ 1 } }{ +^{ 20 } }{ C_{ 2 } }+......{ +^{ 20 } }{ C_{ 10 } } \\ 0{ =^{ 20 } }{ C_{ 0 } }{ -^{ 20 } }{ C_{ 1 } }{ +^{ 20 } }{ C_{ 2 } }+{ ........^{ 20 } }{ C_{ 10 } } \\ =2\left[ { ^{ 20 }{ C_{ 0 } }{ -^{ 20 } }{ C_{ 1 } }+{ { ..... }^{ 20 } }{ C_{ 9 } } } \right] { +^{ 20 } }{ C_{ 10 } } \\ ={ \frac { 1 }{ 2 } ^{ 20 } }{ C_{ 10 } } \end{array}$$

Hence, this is the answer.
Question 5
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Let the $$n^{th}$$ terms of a series be given by
$$t_n$$ = $$\dfrac{n^2 -n -2}{n^2 + 3n}$$ , n 3 .
The product $$t_3$$, $$t_4$$,.........$$t_{50}$$ equals-

A
$$\dfrac{1}{5^2.7.13.53}$$

B
$$\dfrac{1}{5.7^2.12.53}$$

C
$$\dfrac{1}{5^2.7.12.51}$$

D
$$\dfrac{1}{5.7^2.13.53}$$
Question 6
1 / -0

The sum of the series 11.2−12.3+13.4 is equal to

A
$$log_{e}2-1$$

B
$$log_{e}2$$

C
$$log_{e}\left ( \dfrac{4}{e} \right )$$

D
$$2log_{e}2$$

Solution

Given,
$$\frac{1}{1 \cdot 2}-\frac{1}{2 \cdot 3}+\frac{1}{3+4}-\frac{1}{4 \cdot 5}+\cdots$$

We can write it as,
$$\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)+-$$
$$\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+1+\left(-\frac{1}{2}+\frac{1}{3} \frac{1}{4}+\frac{1}{5}\right)\right.$$

because we know,
$$\log (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4} \cdot \cdot \cdot$$

Putting x = 1
We get,
$$\log (2)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5} \ldots \ldots$$

Putting the value of log 2 in the above equation,
$$\Rightarrow \quad \log 2+\log 2-1$$
$$\Rightarrow \quad 2 \log 2-1=\log (2)^{2}-\log e$$
$$= \log {\left(\frac{4}{e}\right)}$$
Option C
Question 7
1 / -0

Match the statements in List 1 with those in List 2
Let $$\alpha, \beta, \gamma$$ be three numbers such that $$\dfrac {1}{\alpha} + \dfrac {1}{\beta} + \dfrac {1}{\gamma} = \dfrac {1}{2}, \dfrac {1}{\alpha^{2}} + \dfrac {1}{\beta^{2}} + \dfrac {1}{\gamma^{2}} = \dfrac {9}{4}$$ and $$\alpha + \beta + \gamma = 2$$, then
List 1 List 2
A. $$\alpha \beta \gamma$$ 1. $$6$$
B. $$\sum \alpha \beta$$ 2. $$-8$$
C. $$\sum \alpha^{2}$$ 3. $$-2$$
D. $$\sum \alpha^{3}$$ 4. $$-1$$

A
$$A-3,B-4,C-1,D-2$$

B
$$A-3,B-1,C-4,D-2$$

C
$$A-4,B-3,C-1,D-2$$

D
$$A-3,B-4,C-2,D-1$$

Solution

(a) $$\left (\sum \dfrac {1}{\alpha}\right )^{2} = \sum \dfrac {1}{\alpha^{2}} + 2\sum = \dfrac {1}{\alpha \beta}$$
or $$\left (\dfrac {1}{2}\right )^{2} = \dfrac {9}{4} + 2\dfrac {(\alpha + \beta + \gamma)}{\alpha \beta \gamma}$$
or $$\dfrac {1}{4} - \dfrac {9}{4} = \dfrac {2.2}{\alpha \beta \gamma} \Rightarrow \alpha \beta \gamma = -2$$
$$\therefore (a)\rightarrow (r)$$
(b) $$\dfrac {1}{\alpha} + \dfrac {1}{\beta} + \dfrac {1}{\gamma} = \dfrac {1}{2} \Rightarrow \dfrac {\sum \alpha \beta}{\alpha \beta \gamma} = \dfrac {1}{2}$$
$$\therefore \dfrac {\sum \alpha \beta}{-2} = \dfrac {1}{2}$$ by $$(a)\Rightarrow \sum \alpha \beta = -1$$
$$\therefore (b)\rightarrow (s)$$
(c) $$\sum \alpha^{2} = (\sum \alpha)^{2} - 2\sum \alpha \beta$$
$$= (2)^{2} - 2(-1) = 6$$ by $$(b)$$
(d) $$\alpha^{3} + \beta^{3} + \gamma^{3} - 3\alpha \beta \gamma = (\alpha + \beta + \gamma)$$
$$(\alpha^{2} + \beta^{2} + \gamma^{2} - \sum \alpha \beta)$$
$$\therefore \sum \alpha^{3} - 3(-2) = 2[-2 - (-1)] =-2$$
$$\therefore \sum \alpha^{3} = -6 - 2 = -8$$.
Question 8
1 / -0

$$1-\dfrac { 3 }{ 2 } +\dfrac { 5 }{ 4 } -\dfrac { 7 }{ 8 } +...\dfrac { 3 }{ 2 } +\dfrac { 5 }{ 4 } -\dfrac { 7 }{ 8 } +...$$

A
$$2/9$$

B
$$2/3$$

C
$$-2/9$$

D
$$9/2$$

Solution
Question 9
1 / -0

If $${\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ......... + {C_n}{x^n},n \in N$$. Then find the value of $$\displaystyle C_0^2 + \frac{{C_1^2}}{2} + {2^3}\frac{{{C_2}}}{3} + {2^4}\frac{{{C_3}}}{4} + ...... + {2^{n }}\frac{{{C_n}}}{{n + 1}} $$

A
$$ \dfrac{{{3^{n - 1}} - 1}}{{n + 1}}$$

B
$$ \dfrac{{{3^{n }} - 1}}{{n + 1}}$$

C
$$ \dfrac{{{3^{n + 1}} - 1}}{{n + 1}}$$

D
$$ \dfrac{{{3^{n + 1}} - 1}}{{n - 1}}$$

Solution
Question 10
1 / -0

The value of $$\dfrac { ^{ a }{ C }_{ 9 } }{ n } +\dfrac { { ^{ n }C_{ 1 } } }{ n+1 } +\dfrac { ^{ n }{ C }_{ 2 } }{ n+2 } +.....+\dfrac { ^{ n }{ C }_{ H } }{ 2n } $$ is equal

A
$$\int _{ 0 }^{ 1 }{ { x }^{ n-1 }\left( 1-x \right) ^{ n }dx } $$

B
$$\int _{ l }^{ 2 }{ { x }^{ n-1 }\left( 1-x \right) ^{ n-1 }dx } $$

C
$$\int _{ l }^{ 2 }{ { x }^{ n-1 }\left( 1-x \right) ^{ n }dx } $$

D
$$\int _{ 0 }^{ 2 }{ { x }^{ n-1 }\left( 1-x \right) ^{ n }{ x }^{ n-1 }dx } $$

Solution