Sequences and Series Test

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Sequences and Series Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

If j, k, and n are consecutive integers such that $$0 < j < k < n$$ and the units (ones) digit of the product jn is 9, what is the units digit of k ?

A
0

B
1

C
2

D
3

E
4

Solution

There are only a few ways you can make a product have a units digit of $$9$$. Either both numbers you’re multiplying have to be $$3$$, or one has to be $$1$$ and one has to be $$9$$. Since we’re dealing with consecutive integers, $$j$$ and $$n$$ can’t both end in $$3$$, so they’re going to have to end in $$1$$ and $$9$$.
It’s important to remember that although we only really care about the units digits in this problem, we’re dealing with numbers that might (or in fact, must) be $$2$$ or more digits. That’s why you can have $$k's$$ units digit be $$0$$.
Say $$j= 19$$, $$k= 20$$, and $$n= 21$$. Then $$jn = (19)(21) = 399$$ (units digit is $$9$$), and the units digit of $$k = 0$$.
Hence option A is correct.
Question 2
1 / -0

The sum of the series $$\sum _{ n=1 }^{ \infty }{ \sin { \left( \cfrac { n!\pi }{ 720 } \right) } } $$ is

A
$$\sin { \left( \cfrac { \pi }{ 180 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } +\sin { \left( \cfrac { \pi }{ 540 } \right) } $$

B
$$\sin { \left( \cfrac { \pi }{ 6 } \right) } +\sin { \left( \cfrac { \pi }{ 30 } \right) } +\sin { \left( \cfrac { \pi }{ 120 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } $$

C
$$\sin { \left( \cfrac { \pi }{ 6 } \right) } +\sin { \left( \cfrac { \pi }{ 30 } \right) } +\sin { \left( \cfrac { \pi }{ 120 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } +\sin { \left( \cfrac { \pi }{ 720 } \right) } $$

D
$$\sin { \left( \cfrac { \pi }{ 180 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } $$

Solution

$$=\sum_{n=1}^{\infty} sin(\dfrac{n!\pi}{6!})$$
$$=sin(\dfrac{1!\pi}{6!})+sin(\dfrac{2!\pi}{6!})+sin(\dfrac{3!\pi}{6!})+...+sin(\dfrac{6!\pi}{6!})+sin(\dfrac{7!\pi}{6!})+....$$
$$=sin(\dfrac{\pi}{720})+sin(\dfrac{\pi}{360})+sin(\dfrac{\pi}{120})+...+sin(\pi)+sin(7\pi)+....$$
$$=sin(\dfrac{\pi}{720})+sin(\dfrac{\pi}{360})+sin(\dfrac{\pi}{120})+sin(\dfrac{\pi}{30})+sin(\dfrac{\pi}{6})$$ [$$\because sin(n\pi)=0$$]
Question 3
1 / -0

$$\displaystyle\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots\cdots+\frac{1}{n(n+1)}=$$

A
$$\displaystyle\frac{1}{n+1}$$

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

C
$$\displaystyle\frac{n+3}{n(n+1)}$$

D
$$\displaystyle\frac{n}{n+1}$$

Solution

$$S_n=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{n(n+1)}=$$

The general term of this series can be written as

$$T_n=\displaystyle\sum^n_{n=1}\dfrac{1}{n(n+1)}$$

$$S_n=\displaystyle\sum^n_{n=1}\dfrac{1}{n(n+1)}$$

$$S_n=\displaystyle\sum^n_{n=1}\dfrac{(n+1)-(n)}{n(n+1)}$$

$$S_n=\displaystyle\sum^n_{n=1}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$$

on putting the values of n

$$S_n=\dfrac{1}{1}-\dfrac{1}{2}$$

$$+\dfrac{1}{2}-\dfrac{1}{3}$$
.
.
.
$$+\dfrac{1}{n-1}-\dfrac{1}{n}$$

$$+\dfrac{1}{n}-\dfrac{1}{n+1}$$

$$S_n=1-\dfrac{1}{n+1}\\$$
$$S_n=\dfrac{n+1-1}{n+1}\\$$
$$S_n=\dfrac{n}{n+1}$$.
Question 4
1 / -0

The variance of the series
$$a,a+d,a+2d,.....a+(2n-1)d,a+2nd$$ is

A
$$\cfrac { n(n+1) }{ 2 } { d }^{ 2 }$$

B
$$\cfrac { n(n-1) }{ 6 } { d }^{ 2 }$$

C
$$\cfrac { n(n+1) }{ 6 } { d }^{ 2 }$$

D
$$\cfrac { n(n+1) }{ 3 } { d }^{ 2 }$$

Solution

The mean of the given series $$a+nd$$
$$\therefore$$ Variance $$=\cfrac { 1 }{ 2n+1 } \sum _{ r=0 }^{ 2n }{ { \left\{ (a+rd)-(a+nd) \right\} }^{ 2 } } $$
$$\Rightarrow $$ Variance $$\quad =\cfrac { { d }^{ 2 } }{ 2n+1 } \sum _{ r=0 }^{ 2n }{ { \left| r-n \right| }^{ 2 } } $$
$$\Rightarrow $$ Variance $$=\cfrac { { d }^{ 2 } }{ 2n+1 } \sum _{ r=0 }^{ 2n }{ { (n-r) }^{ 2 } } $$
$$\Rightarrow $$ Variance $$=\cfrac { { 2d }^{ 2 } }{ 2n+1 } \sum _{ r=1 }^{ n }{ { r }^{ 2 } } $$
$$\Rightarrow $$ Variance $$=\cfrac { { 2d }^{ 2 } }{ 2n+1 } \times \cfrac { n(n+1)(2n+1) }{ 6 } \quad =\cfrac { n(n+1) }{ 3 } { d }^{ 2 }$$
Question 5
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It is given that $$\sum_{r = 1}^{\infty} \dfrac{1}{(2 r - 1)^2} = \dfrac{\pi^2}{8}$$, then $$\sum_{r = 1}^{\infty} \dfrac{1}{r^2}$$ is equal to

A
$$\dfrac{\pi^2}{24}$$

B
$$\dfrac{\pi^2}{3}$$

C
$$\dfrac{\pi^2}{6}$$

D
None of these

Solution

$$^{\infty } _{r=1}\Sigma \dfrac{1}{\left ( 2r-1 \right )^{2}}=\dfrac{\pi^{2} }{8}$$

$$\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+\dfrac{1}{7^{2}}+........=\dfrac{\pi^{2} }{8}$$

$$I= \dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{5}+\dfrac{1}{6^{2}}+\dfrac{1}{7^{2}}$$
$$I=\left ( \dfrac{1}{2^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{6^{2}}....... \right )+\left ( \dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+........ \right )$$

$$=\dfrac{1}{2^{2}}\left ( \dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}.......... \right )+\dfrac{\pi^{2} }{8}$$
$$\dfrac{4I-I}{4}=\dfrac{\pi^{2} }{8}$$
$$\dfrac{4I-I}{4}=\dfrac{\pi^{2} }{8}$$
$$\dfrac{3I}{4}=\dfrac{\pi ^{2}}{8}$$
$$I=\dfrac{\pi ^{2}}{6}$$
Question 6
1 / -0

Let $$\{a_n\}$$ be a sequence of numbers satisfying the relation $$(3-a_{n+1})(6+a_n)=18$$ for all $$n\ge 0$$ and $$a_0=3$$. Then
$$\displaystyle \underset{n\rightarrow \infty}{lim}\dfrac{1}{2^{n+2}}\sum_{j=0}^{n}\dfrac{1}{a_j}$$

A
$$\dfrac{1}{18}$$

B
$$\dfrac{1}{6}$$

C
$$\dfrac{1}{4}$$

D
$$\dfrac{1}{3}$$

Solution

Given $$\left( 3-{ a }_{ n+1 } \right) \left( 6+{ a }_{ n } \right) =18$$
$$3-{ a }_{ n+1 }= \cfrac { 18 }{ 6+{ a }_{ n } } $$
$${ a }_{ n+1 }=3-\cfrac { 18 }{ 6+{ a }_{ n } } = \cfrac { 3{ a }_{ n } }{ 6+{ a }_{ n } }$$
$$a_0=3,a_1=1,a_2=\cfrac{3}{7},a_3=\cfrac{1}{5},a_4=\cfrac{3}{31},a_5=\cfrac{1}{21}$$
Let
$$S={ \sum }_{ j=0 }^{ n }\cfrac { 1 }{ { a }_{ j } } =\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 1 } +\cfrac { 7 }{ 3 } +\cfrac { 8 }{ 1 } +\cfrac { 31 }{ 3 } +\cfrac { 21 }{ 1 } +...\\ \quad S=\cfrac { 1 }{ 3 } +1+\cfrac { 7 }{ 3 } +\cfrac { 5 }{ 1 } +\cfrac { 31 }{ 3 } +\cfrac { 21 }{ 1 } +...+{ a' }_{ n }+{ a' }_{ n+1 }\\ -\underline { \;S=\;\;\;\quad \cfrac { 1 }{ 3 } +1+\cfrac { 7 }{ 3 } +\cfrac { 5 }{ 1 } +\cfrac { 31 }{ 3 } +...+{ a' }_{ n }+{ a' }_{ n+1 } } \\ \quad\; 0=\;\cfrac { 1 }{ 3 } +\cfrac { 2 }{ 3 } +\cfrac { 4 }{ 3 } +\cfrac { 8 }{ 3 } +\cfrac { 16 }{ 3 } +\cfrac { 32 }{ 3 } +....-{ a' }_{ n+1 }\\ { a' }_{ n+1 }=\cfrac { 1 }{ 3 } \left( 1+2+4+8+16+32+.... \right) $$

This is a GP with $$r=2 \quad S_n=\cfrac{a(r^n-1)}{(r-1)}$$
$${ a }_{ n }=\cfrac { 1 }{ 3 } \cfrac { (1){ (2) }^{ n+1 }-1 }{ (2-1) } =\cfrac { 1 }{ 3 } ({ 2 }^{ n+1 }-1)$$
$${ \sum }_{ i=0 }^{ n }{ a' }_{ n }$$$$=\cfrac { 1 }{ 3 } \left[ { \Sigma }_{ i=0 }^{ n }{ 2 }^{ n+1 }-{ \Sigma }_{ i=0 }^{ n }n \right] \\ =\cfrac { 1 }{ 3 } \left[ 2\cfrac { ({ 2 }^{ n }-1) }{ 2-1 } -n \right] \\ =\cfrac { 1 }{ 3 } \left( { 2 }^{ n+1 }-2-n \right) $$
$$\underset { n\rightarrow \infty }{ \lim } \cfrac { 1 }{ 2n-12 } { \sum }_{ i=0 }^{ n }\cfrac { 1 }{ { a }_{ i } } =\underset { n\rightarrow \infty }{ \lim } \cfrac { 1 }{ 6 } \left[ \cfrac { { 2 }^{ n+1 }-2-n }{ { 2 }^{ n+1 } } \right] \\ =\cfrac { 1 }{ 6 } \underset { n\rightarrow \infty }{ \lim } \left[ 1-\cfrac { 2 }{ { 2 }^{ n+1 } } -\cfrac { n }{ { 2 }^{ n+1 } } \right] =\cfrac { 1 }{ 6 } $$
Question 7
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If $$\displaystyle \sum _{ r =1 }^{n}{ t_r = \frac{n(n+1)(n+2)(n+3)}{8} }$$, then $$\displaystyle \sum _{r = 1}^{n}{ \frac{1}{t_r}} $$ equals

A
$$\displaystyle - \left( \frac{1}{(n+1)(n+2)} - \frac{1}{2} \right) $$

B
$$\displaystyle \left( \frac{1}{(n+1)(n+2)} - \frac{1}{2} \right) $$

C
$$\displaystyle \left( \frac{1}{(n+1)(n+2)} + \frac{1}{2} \right) $$

D
$$\displaystyle \left( \frac{1}{(n-1)(n-2)} + \frac{1}{2} \right) $$

Solution

Let $${ S }_{ n }=\sum _{ r=1 }^{ n }{ { t }_{ r } } $$

ie, $${ t }_{ n }={ S }_{ n }-{ S }_{ n-1 }=\dfrac { n(n+1)(n+2)(n+3) }{ 8 } -\dfrac { (n-1)n(n+1)(n+2) }{ 8 } $$
$$=\dfrac { n(n+1)(n+2)(n+3-(n-1)) }{ 8 } $$
$$=\dfrac { n(n+1)(n+2) }{ 2 } $$

Let, $$\dfrac { 1 }{ { t }_{ n } } =\dfrac { 2 }{ n(n+1)(n+2) } =\dfrac { A }{ n } +\dfrac { B }{ n+1 } +\dfrac { C }{ n+2 } $$

$$\Rightarrow \dfrac { A }{ n } +\dfrac { B }{ n+1 } +\dfrac { C }{ n+2 } =\dfrac { A(n+1)+Bn }{ n(n+1) } +\dfrac { C }{ n+2 } =\dfrac { ((A+B)n+A)(n+2)+Cn(n+1) }{ n(n+1)(n+2) } $$

$$\Rightarrow \dfrac { (A+B+C){ n }^{ 2 }+(3A+2B+C)n+2A }{ n(n+1)(n+2) } =\dfrac { 2 }{ n(n+1)(n+2) } $$

ie,$$A=1,B=(-2),C=1$$
Therefore, $$ \sum _{ r=1 }^{ n }{ \dfrac { 1 }{ { t }_{ r } } =\sum _{ r=1 }^{ n }{ (\dfrac { 1 }{ n } -\dfrac { 2 }{ n+1 } +\dfrac { 1 }{ n+2 } ) } } =\left (\dfrac { 1 }{ 1 } +\boxed { -\dfrac { 2 }{ 2 } \\ +\dfrac { 1 }{ 2 } } +\boxed { +\dfrac { 1 }{ 3 } \\ -\dfrac { 2 }{ 3 } +\\ +\dfrac { 1 }{ 3 } } +\boxed { +\dfrac { 1 }{ 4 } \\ -\dfrac { 2 }{ 4 } \\ +\dfrac { 1 }{ 4 } } +\boxed { ... } +\boxed { +\dfrac { 1 }{ n } \\ -\dfrac { 2 }{ n } \\ +\dfrac { 1 }{ n } } +\boxed { +\dfrac { 1 }{ n+1 } \\ -\dfrac { 2 }{ n+1 } } +\dfrac { 1 }{ n+2 }\right )$$

$$\Rightarrow \sum _{ r=1 }^{ n }{ \dfrac { 1 }{ { t }_{ r } } } =\dfrac { 1 }{ 2 } -\dfrac { 1 }{ n+1 } +\dfrac { 1 }{ n+2 } =\dfrac { 1 }{ 2 } -\dfrac { 1 }{ (n+1)(n+2) } $$
$$=-\left(\dfrac {1}{(n+1)(n+2)}-\dfrac {1}{2}\right)$$

So, option A is correct
Question 8
1 / -0

Let $$A=16{-4}+2^{-4}+3^{-4}+4^{-4}+...$$ and $$B=1^{-4}+3^{-4}+5^{-4}+7^{-4}+...$$. The ratio $$\dfrac{A}{B}$$ in the lowest form is

A
$$\dfrac{16}{15}$$

B
$$\dfrac{15}{14}$$

C
$$\dfrac{15}{16}$$

D
$$\dfrac{13}{12}$$

Solution

$$A=16-4+{ 2 }^{ -4 }+{ 3 }^{ -4 }+{ 4 }^{ -4 }+....$$
$$A=\left( 11+{ 1 }^{ -4 }+{ 2 }^{ -4 }+.... \right) $$
$$\displaystyle \sum _{ n=1 }^{ \infty }{ \cfrac { 1 }{ { n }^{ 4 } } } =\cfrac { {\pi}^{ 4 } }{ 90 } $$
$$A=11+\cfrac { { \pi }^{ 4 } }{ 90 } $$
$$B={ 1 }^{ -4 }+{ 3 }^{ -4 }+{ 5 }^{ -4 }+....$$
$$=\left( { 1 }^{ -4 }+{ 2 }^{ -4 }+{ 3 }^{ -4 }+....\infty \right) -\left( { 2 }^{ -4 }+{ 4 }^{ -4 }+{ 6 }^{ -4 }+.... \right) $$
$$B=\displaystyle \cfrac { { \pi }^{ 4 } }{ 90 } -\sum _{ n=1 }^{ \infty }{ \cfrac { 1 }{ { \left( 2n \right) }^{ 4 } } } $$
$$=\displaystyle \cfrac { { \pi }^{ 4 } }{ 90 } -\cfrac { 1 }{ 16 } \sum _{ n=1 }^{ \infty }{ \cfrac { 1 }{ { n }^{ 4 } } } =\cfrac { { \pi }^{ 4 } }{ 90 } -\cfrac { { \pi }^{ 4 } }{ 90\times 16 } $$
$$=\left( \cfrac { 15 }{ 16 } \right) \cfrac { { \pi }^{ 4 } }{ 90 } $$
$$\cfrac {A}{B} =\cfrac { 11+\cfrac { { \pi }^{ 4 } }{ 90 } }{ \left( \cfrac { 15 }{ 16 } \right) \cfrac { { \pi }^{ 4 } }{ 90 } } =\left( \cfrac { 16 }{ 15 } \right) \left[ \cfrac { 90\times 11 }{ { \pi }^{ 4 } } +1 \right] $$
Lowest form will be $$\cfrac { 16 }{ 15 } $$
Question 9
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Which of the following options will continue the given series?
$$29$$, $$34$$, $$32$$, $$37$$, $$35$$, ?

A
$$36$$

B
$$39$$

C
$$40$$

D
$$42$$
Question 10
1 / -0

The value of $$S=\sqrt { 1+\cfrac { 1 }{ { 1 }^{ 2 } } +\cfrac { 1 }{ { 2 }^{ 2 } } } +\sqrt { 1++\cfrac { 1 }{ { 2 }^{ 2 } } +\cfrac { 1 }{ { 3 }^{ 2 } } } +...+\sqrt { 1++\cfrac { 1 }{ { (2014) }^{ 2 } } +\cfrac { 1 }{ { (2015) }^{ 2 } } } $$ is

A
$$2015$$

B
$$2015-\cfrac { 1 }{ 2015 } $$

C
$$2016-\cfrac { 1 }{ 2016 } $$

D
$$2014-\cfrac { 1 }{ 2014 } $$

Solution

The first term $$\sqrt{1+\dfrac1{1^2}+\dfrac1{2^2}}=\sqrt{1+1+\dfrac14}=\sqrt{\dfrac94}=\dfrac32=1+\dfrac12=1+1-\dfrac12$$
The second term $$\sqrt{1+\dfrac1{2^2}+\dfrac1{3^2}}=\sqrt{1+\dfrac14+\dfrac19}=\sqrt{\dfrac{49}{36}}=\dfrac76=1+\dfrac16=1+\dfrac12-\dfrac13$$
The Third term similarily can be written as $$1+\dfrac13-\dfrac14$$
Hence, $$S=\left (1+1-\dfrac12\right)+\left (1+\dfrac12-\dfrac13\right)+...+\left (1+\dfrac1{2014}-\dfrac1{2015}\right)=2015-\dfrac1{2015}$$

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Sequences and Series Test - 54

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Sequences and Series Test - 54

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Question 1

0

1

2

3

4

Solution

Question 2

$$\sin { \left( \cfrac { \pi }{ 180 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } +\sin { \left( \cfrac { \pi }{ 540 } \right) } $$

$$\sin { \left( \cfrac { \pi }{ 6 } \right) } +\sin { \left( \cfrac { \pi }{ 30 } \right) } +\sin { \left( \cfrac { \pi }{ 120 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } $$

$$\sin { \left( \cfrac { \pi }{ 6 } \right) } +\sin { \left( \cfrac { \pi }{ 30 } \right) } +\sin { \left( \cfrac { \pi }{ 120 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } +\sin { \left( \cfrac { \pi }{ 720 } \right) } $$

$$\sin { \left( \cfrac { \pi }{ 180 } \right) } +\sin { \left( \cfrac { \pi }{ 360 } \right) } $$

Solution

Question 3

$$\displaystyle\frac{1}{n+1}$$

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

$$\displaystyle\frac{n+3}{n(n+1)}$$

$$\displaystyle\frac{n}{n+1}$$

Solution

Question 4

$$\cfrac { n(n+1) }{ 2 } { d }^{ 2 }$$

$$\cfrac { n(n-1) }{ 6 } { d }^{ 2 }$$

$$\cfrac { n(n+1) }{ 6 } { d }^{ 2 }$$

$$\cfrac { n(n+1) }{ 3 } { d }^{ 2 }$$

Solution

Question 5

$$\dfrac{\pi^2}{24}$$

$$\dfrac{\pi^2}{3}$$

$$\dfrac{\pi^2}{6}$$

None of these

Solution

Question 6

$$\dfrac{1}{18}$$

$$\dfrac{1}{6}$$

$$\dfrac{1}{4}$$

$$\dfrac{1}{3}$$

Solution

Question 7

$$\displaystyle - \left( \frac{1}{(n+1)(n+2)} - \frac{1}{2} \right) $$

$$\displaystyle \left( \frac{1}{(n+1)(n+2)} - \frac{1}{2} \right) $$

$$\displaystyle \left( \frac{1}{(n+1)(n+2)} + \frac{1}{2} \right) $$

$$\displaystyle \left( \frac{1}{(n-1)(n-2)} + \frac{1}{2} \right) $$

Solution

Question 8

$$\dfrac{16}{15}$$

$$\dfrac{15}{14}$$

$$\dfrac{15}{16}$$

$$\dfrac{13}{12}$$

Solution

Question 9

$$36$$

$$39$$

$$40$$

$$42$$

Question 10

$$2015$$

$$2015-\cfrac { 1 }{ 2015 } $$

$$2016-\cfrac { 1 }{ 2016 } $$

$$2014-\cfrac { 1 }{ 2014 } $$

Solution

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