Sequences and Series Test

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Sequences and Series Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

The sum of the series $$1+\dfrac{\log_e x}{1!}+\dfrac{(\log_e x)^2}{2!}+........$$ is

A
$$x$$

B
$$x^2$$

C
$$x^3$$

D
none of these

Solution

We've,
$$1+\dfrac{\log_e x}{1!}+\dfrac{(\log_e x)^2}{2!}+........$$
$$=e^{\log_ex}$$[ Since $$e^x=1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+........$$]
$$=x$$.
Question 2
1 / -0

If x<1, then $$\displaystyle \frac { 1 }{ 1+x } +\frac { 2x }{ 1+{ x }^{ 2 } } +\frac { { 4x }^{ 3 } }{ 1+{ x }^{ 4 } } +.........\infty =$$

A
$$x$$

B
$$\frac { 1 }{ 1+x } $$

C
$$\frac { 1 }{ 1-x } $$

D
$$\frac { 1 }{ x } $$

Solution
Question 3
1 / -0

Find the sum of
$$1+\frac { 1 }{ 4 } +\frac { 1.3 }{ 4.8 } +\frac { 1.3.5 }{ 4.8.12 } +...\infty $$.

A
$$2\sqrt { 2 } $$

B
$$2\sqrt { 3 } $$

C
$$\sqrt { 2 } $$

D
$$\sqrt { \frac { 1 }{ 2 } } $$

Solution

solution
$$1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\cdots \cdot$$

$$(1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+ \cdots$$

' $$n$$ ' can be fractional, negative also.

$$n x=\frac{1}{4} \quad \frac{n(n-1)}{2} x^{2}=\frac{3}{32}-(11) \\$$

$$n^{2} x^{2}=\frac{1}{16} \\$$

$$x^{2}=\frac{1}{16 n^{2}}-(1)$$

Put value of $$x^{2}$$ from equation (1) in equation (11)

$$\frac{n(n-1)}{2} \frac{1}{16 n^{2}}=\frac{3}{32} \Rightarrow \frac{n^{2}-n}{n^{2}}=3$$

$$\Rightarrow n^{2}-n=3 n^{2} \Rightarrow 2 n^{2}+n=0$$

$$\Rightarrow n(2 n+1)=0$$

$$n=0$$ which is not possible here so $$n=-\frac{1}{2} \quad x=-\frac{1}{2}$$

$$=\frac{1}{\sqrt{2}}$$

Answer : option $$(D)$$
Question 4
1 / -0

The sum of the series 1+2(1 +1/n)+ 3$${ \left( 1+1/n \right) }^{ 2 }+....\infty \quad is\quad given\quad by$$

A
$${ n }^{ 2 }\quad +\quad 1$$

B
$$n\left( n+1 \right) $$

C
$$n{ (1+1/n) }^{ 2 }$$

D
$${ n }^{ 2 }$$

Solution

Given,

$$S_m=1+2\left ( 1+\dfrac{1}{n} \right )+.........+\infty $$

$$\Rightarrow S_m\left ( 1+\dfrac{1}{n} \right )=\left ( 1+\dfrac{1}{n} \right )+2\left ( 1+\dfrac{1}{n} \right )^2+.....+\infty $$

$$S_m-S_m\left ( 1+\dfrac{1}{n} \right )=-\dfrac{S_m}{n}$$

$$-\dfrac{S_m}{n}=1+\left ( 1+\dfrac{1}{n} \right )+\left ( 1+\dfrac{1}{n} \right )^2+.........+\infty $$

$$-\dfrac{S_m}{n}=\dfrac {1}{1-\left ( 1+\dfrac{1}{n} \right )}=-n$$

$$S_m=n^2$$
Question 5
1 / -0

The $${ 7 }^{ th }$$ term of the series$$1+\dfrac { 1 }{ \left( 1+3 \right) } { \left( 1+2 \right) }^{ 2 }\dfrac { 1 }{ \left( 1+3+5 \right) } { \left( 1+2+3 \right) }^{ 2 }+.....,$$ is equal to ________.

A
16

B
18

C
20

D
22

Solution
Question 6
1 / -0

The sum of the first n terms of the series $${ 1 }^{ 2 }+2.{ 2 }^{ 2 }+{ 3 }^{ 2 }+2.{ 4 }^{ 2 }+{ 5 }^{ 2 }+2.{ 6 }^{ 2 }+...$$ is $$\dfrac { n{ \left( n+1 \right) }^{ 2 } }{ 2 } $$ when n is even. When n is odd the sum is

A
$$\dfrac { 3n\left( n+1 \right) }{ 2 } $$

B
$$\dfrac { { n }^{ 2 }\left( n+1 \right) }{ 2 } $$

C
$$\dfrac { { n\left( n+1 \right) }^{ 2 } }{ 4 } $$

D
$${ \left[ \dfrac { n\left( n+1 \right) }{ 2 } \right] }^{ 2 }$$

Solution
Question 7
1 / -0

The sum of the series $$ 1 + \frac { 1 + 2 } { 2 } + \frac { 1 + 2 + 3 } { 3 } + \ldots $$ to n terms is

A
$$

\frac { n ( n + 1 ) } { 2 }

$$

B
$$

\frac { n ( n + 3 ) } { 4 }

$$

C
$$

\frac { n ( n + 1 ) + n } { 4 }

$$

D
$$

\frac { n ( n - 1 ) } { 2 }

$$

Solution
Question 8
1 / -0

If $$\sum _{ k=2 }^{ n }{ cos{ }^{ -1 }(\frac { 1+\sqrt { (k-1)(k+2)(k+1)k } }{ k(k+1) } } )=\frac { 120\pi }{ \lambda }$$,then

A
number of even divisors of $$\lambda$$ is 24

B
Sum of proper divisors of $$\lambda$$ is 2418.

C
Sum of proper divisors of $$\lambda$$ is 197.

D
number of ways in which $$\lambda$$ can be expressed as product of two co-prime factors is 4
Question 9
1 / -0

Use geometric series to express $$0.555...=0.\overline 5$$ as a rational number.

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

B
$$\dfrac{5}{9}$$

C
$$\dfrac{5}{99}$$

D
None of these

Solution

$$0.\overline 5=0.5555...$$

$$0.\overline 5=0.5 +0.05 +0.005 +... \infty$$

$$0.\overline 5=\dfrac{5}{10}+\dfrac{5}{10^2}+\dfrac{5}{10^3}+... \infty$$

This is an infinite $$G.P$$ with $$a=\dfrac {5}{10}$$ and $$r=\dfrac {1}{10}$$

$$0.\overline 5=\dfrac{\left(\dfrac {5}{10}\right)}{1-\left(\dfrac {1}{10}\right)}=\dfrac{5}{9}$$
Question 10
1 / -0

If $$L=\sum _{ r=7 }^{ 2400 }{ \log _{ 7 }{ \left( \frac { r+1 }{ r } \right) } } ,M=\prod _{ r=2 }^{ 1023 }{ \log _{ r }{ (r+1) } } $$ and $$N=\sum _{ r=2 }^{ 2011 }{ \left( \frac { 1 }{ \log _{ r }{ p } } \right) } $$ where p=$$(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...........\cdot 2011)$$, then

A
$$L+M=13$$

B
$$M^2+N^2=101$$

C
$$L-M+N=6$$

D
$$LMN=30$$

Solution

$$\text { Solve: } \\$$

$$\text { Given, } \\$$

$$\qquad L=\sum_{r=7}^{2400} \log _{7}\left(\frac{r+1}{r}\right) \\$$

$$\left.\Rightarrow L=\log _{7}\left(\frac{8}{7}\right)+\log _{7}\left(\frac{9}{8}\right)+\log _{7}\left(\frac{10}{9}\right)+-\cdots+\log _{7} (\frac{2401}{2400}\right)\\$$

$$\text { We know that } \\$$

$$\qquad \log a+\log b=\log a b$$

$$\Rightarrow L=\log _{7}\left(\frac{8}{7}\right) \times\left(\frac{9}{8}\right) \times\left(\frac{10}{9}\right) \times \ldots \times\left(\frac{401}{2400}\right)$$

$$\Rightarrow L=\log _{7}\left(\frac{240}{7}\right)=\log _{7} 7^{3}$$

$$\Rightarrow L=3 \quad-(i) \quad\left\{\log _a {a}^{x}=x\right.$$

$$M =\pi_{r=2}^{1023} \log _{r}(r+1) \\$$

$$\Rightarrow M=\log _{2}(3) \cdot \log _{3} 4 \cdot \log _{4} 5 . . \log _{1023} 1024 \\$$

$$\Rightarrow \quad M=\left(\frac{\log 3}{\log 2}\right) \cdot\left(\frac{\log 4}{\log 3}\right) . \cdots\left(\frac{\log 1024}{\log 1023}\right) $$

$$\left\{\log _{a} b=\frac{\log {b}}{\log a}\right\}$$

$$\Rightarrow M =\frac{\log 1024}{\log 2}=\log _{2} 2^{10} \\ $$

$$\Rightarrow M=10 \quad-(i i) $$

Now, $$N=\sum_{r=2}^{20 \pi}\left(\frac{1}{\log _{r}{p}}\right)$$

Where, $$P=(1.2 .3 .4 .5 \ldots . .2011)$$

$$\Rightarrow P=2011 !$$

$$\Rightarrow N=\log _{p}^{2}+\log _{p}^{3}+\log _{p}^{4}+\ldots+\log _{p}^{2011}$$

$$\Rightarrow N=\log _{p}(2 \cdot 3 \cdot 4 \cdot \ldots 2011)=\log _{p}(2011) !$$

$$\Rightarrow N=\log _{p}^{p}=1 \quad-(i i i)$$

on adding equation (i) and equation (ii)

we get.

$$L+M=3+10=13$$

from eqn. (i), (i) and (iii) we get

$$L=3, M=10$$ and $$N=1$$

$$\Rightarrow \quad M^{2}+N^{2}=100+1=101$$

and $$\quad$$ LMN $$=3 \cdot 10 \cdot 1=30$$

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Sequences and Series Test - 50

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Sequences and Series Test - 50

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SHARING IS CARING

Question 1

$$x$$

$$x^2$$

$$x^3$$

none of these

Solution

Question 2

$$x$$

$$\frac { 1 }{ 1+x } $$

$$\frac { 1 }{ 1-x } $$

$$\frac { 1 }{ x } $$

Solution

Question 3

$$2\sqrt { 2 } $$

$$2\sqrt { 3 } $$

$$\sqrt { 2 } $$

$$\sqrt { \frac { 1 }{ 2 } } $$

Solution

Question 4

$${ n }^{ 2 }\quad +\quad 1$$

$$n\left( n+1 \right) $$

$$n{ (1+1/n) }^{ 2 }$$

$${ n }^{ 2 }$$

Solution

Question 5

16

18

20

22

Solution

Question 6

$$\dfrac { 3n\left( n+1 \right) }{ 2 } $$

$$\dfrac { { n }^{ 2 }\left( n+1 \right) }{ 2 } $$

$$\dfrac { { n\left( n+1 \right) }^{ 2 } }{ 4 } $$

$${ \left[ \dfrac { n\left( n+1 \right) }{ 2 } \right] }^{ 2 }$$

Solution

Question 7

$$\frac { n ( n + 1 ) } { 2 }$$

$$\frac { n ( n + 3 ) } { 4 }$$

$$\frac { n ( n + 1 ) + n } { 4 }$$

$$\frac { n ( n - 1 ) } { 2 }$$

Solution

Question 8

number of even divisors of $$\lambda$$ is 24

Sum of proper divisors of $$\lambda$$ is 2418.

Sum of proper divisors of $$\lambda$$ is 197.

number of ways in which $$\lambda$$ can be expressed as product of two co-prime factors is 4

Question 9

$$\dfrac{1}{5}$$

$$\dfrac{5}{9}$$

$$\dfrac{5}{99}$$

None of these

Solution

Question 10

$$L+M=13$$

$$M^2+N^2=101$$

$$L-M+N=6$$

$$LMN=30$$

Solution

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$$

\frac { n ( n + 1 ) } { 2 }

$$

$$

\frac { n ( n + 3 ) } { 4 }

$$

$$

\frac { n ( n + 1 ) + n } { 4 }

$$

$$

\frac { n ( n - 1 ) } { 2 }

$$