Sequences and Series Test

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Sequences and Series Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

The value of the infinite series $$\cfrac{1^2+2^2}{\underline{|3}} + \cfrac{1^2+2^2+3^2}{\underline{|4}}+\cfrac{1^2+2^2+3^3+4^2}{\underline{|5}}........$$ is

A
$$e$$

B
$$5e$$

C
$$\cfrac{5e}{6} -\cfrac{1}{2}$$

D
$$\cfrac{5e}{6}$$

Solution

$${ T }_{ n }=\cfrac { n(n+1)(2n+1) }{ 6(n+1)! } \\ S={ \sum }_{ i=2 }^{ \infty }\cfrac { n(n+1)(2n+1) }{ 6(n+1)! } ={ \sum }_{ i=2 }^{ \infty }\cfrac { (2n+1) }{ 6(n-1)! } \\ S={ \sum }_{ i=2 }^{ \infty }\cfrac { 2n-2+3 }{ 6(n-1)! } \\ S=\cfrac { 1 }{ 6 } { \sum }_{ i=2 }^{ \infty }\cfrac { 2(n-1) }{ (n-1)! } +\cfrac { 3 }{ 6 } { \sum }_{ i=2 }^{ \infty }\cfrac { 1 }{ (n-1)! } \\ S=\cfrac { 2 }{ 6 } { \sum }_{ i=2 }^{ \infty }\cfrac { 1 }{ (n-2)! } +\cfrac { 1 }{ 2 } { \sum }_{ i=2 }^{ \infty }\cfrac { 1 }{ (n-1)! } \\ S=\cfrac { 1 }{ 3 } e+\cfrac { 1 }{ 2 } (e-1)\\ S=\cfrac { 5e }{ 6 } -\cfrac { 1 }{ 2 } $$
Question 2
1 / -0

Let $$S$$ denote the sum of the infinite series $$1+\cfrac { 8 }{ 2! } +\cfrac { 21 }{ 3! } +\cfrac { 40 }{ 4! } +\cfrac { 65 }{ 5! } +.......$$. Then

A
$$S< 8$$

B
$$S> 12$$

C
$$8< S< 12$$

D
$$S=8$$

Solution

$$\; \;\; S_n=1+8+21+40+65+.....+a_n \\ - \underline{ S_n=\quad \; \;1+\;8+21+40+.....+a_{n-1}+a_n }$$
$$\;\;\;0\;=1+7+13+19+25+...,,,,,,,,.-a_n$$
$${ a }_{ n }=1+7+13+19+25+....\\ { a }_{ n }=\cfrac { n }{ 2 } [2a+(n-1)d]=\cfrac { n }{ 2 } [2+(n-1)6]\\ { a }_{ n }=n(3n-2)\\ S={ \sum }_{ i=1 }^{ \infty }\cfrac { n(3n-2) }{ n! } ={ \sum }_{ i=1 }^{ \infty }\cfrac { (3n-2) }{ (n-1)! } $$
Adding and subtracting $$1$$ in numerator
$$S={ \sum }_{ i=2 }^{ \infty }\cfrac { 3(n-1) }{ (n-1)! } +{ \sum }_{ i=1 }^{ \infty }\cfrac { 1 }{ (n-1)! } \\ S=3{ \sum }_{ i=2 }^{ \infty }\cfrac { 1 }{ (n-2)! } +{ \sum }_{ i=1 }^{ \infty }\cfrac { 1 }{ (n-1)! } \\ S=3e+e\\ S=4e\\ 8<S<12\\ { e }^{ x }=1+\cfrac { x }{ 1! } +\cfrac { { x }^{ 2 } }{ 2! } +\cfrac { { x }^{ 3 } }{ 3! } +\cfrac { { x }^{ 4 } }{ 4! } +....\infty \\ e=1+\cfrac { 1 }{ 1! } +\cfrac { 1 }{ 2! } +\cfrac { 1 }{ 3! } +\cfrac { 1 }{ 4! } +....\infty $$
Question 3
1 / -0

$$x=1+\frac{1}{2\times \underline{|1}}+\frac{1}{4\times \underline{|2}}+\frac{1}{8\times \underline{|3}}$$

A
$${e}^{1/2}$$

B
$$e^2$$

C
e

D
$$\frac{1}{e}$$

Solution

We know,
$$e^{x}=1+\dfrac{x}{1!}+\dfrac{x^{2}}{2!}+\dfrac{x^{3}}{3!}+...$$
put $$x=\dfrac{1}{2}$$
$$\therefore e^{\frac{1}{2}}=1+\dfrac{1}{2\times1!}+\dfrac{1}{4\times2!}+\dfrac{1}{8\times3!}+...$$
$$\therefore Ans=\sqrt{e}$$
Question 4
1 / -0

In the series $$1+3+6+10+......,$$ find the $$n^{th}$$ term:

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

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

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

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

Solution

$$1=1$$, here $$n=1\Rightarrow \dfrac {1(1+1)}{2}=1$$
$$1+2=3$$, here $$n=2\Rightarrow \dfrac {2(2+1)}{2}=3$$

$$1+2+3=6$$, here $$n=3\Rightarrow \dfrac {3(3+1)}{2}=6$$

$$1+2+3+4=10$$, here $$n=4\Rightarrow \dfrac {4(4+1)}{2}=10$$

$${ n }^{ th }$$ term is sum of $$n$$ terms

Therefore, $${ T }_{ n }=\dfrac { n(n+1) }{ 2 } $$
Question 5
1 / -0

Consider an incomplete pyramid of balls on a square base having $$18$$ layers; and having $$13$$ balls on each side of the top layer. Then the total number $$N$$ of balls in that pyramid satisfies

A
$$9000 < N < 10000$$

B
$$8000 < N < 9000$$

C
$$7000 < N < 8000$$

D
$$10000 < N < 12000$$

Solution

The top layer has $$(13\times 13)$$ balls the layer below it will have $$(14\times 14)$$ balls

We have $$18$$ layers

So the total number of balls

$$N=(13\times 13)+(14\times 14)+.......(30\times 30)\\ N={ 13 }^{ 2 }+{ 14 }^{ 2 }+.....{ 30 }^{ 2 }$$

Sum of squares of first $$n$$ natural numbers is $$\dfrac { n(n+1)(2n+1) }{ 6 } \\ $$

$$\therefore N=$$ sum of first $$30$$ $$-$$ sum of first $$12$$

$$N=\dfrac { 30\times 31\times 61 }{ 6 } -\dfrac { 12\times 13\times 25 }{ 6 } \\ N=8805\\ $$

$$\Rightarrow 8000<N<9000$$
Hence, option B is correct.
Question 6
1 / -0

With the help of match-sticks, Zalak prepared a pattern as shown below. When $$97$$ matchsticks are used, the serial number of the figure will be ...........

A
Figure 32

B
Figure 95

C
Figure 49

D
Figure 48

Solution

Pattern associated with the number of matchsticks:
Figure 1: No. of sticks $$=3=3+0$$
Figure 2: No. of sticks $$=5=3+2$$
Figure 3: No. of sticks $$=7=3+2+2$$
Figure 4: No. of sticks $$=9=3+2+2+2$$
----------------------------------------
The rule associated with it can be given as no. of sticks $$=3+2(n-1)$$ where $$n$$ is the number of the figure.
Now there are $$97$$ matchsticks
$$\implies 3+2(n-1)=97$$
$$\implies 3+2n-2=97$$
$$\implies 2n=96$$
$$\implies n=48$$
Hence, serial number of the figure having $$97$$ matchsticks is $$48$$.
Question 7
1 / -0

Find the $$(2n)^{th}$$ term of the series whose $$n^{th}$$ term is $$\dfrac{n^2+1}{n^3}$$:

A
$$\dfrac{n^2+1}{8n^3}$$

B
$$\dfrac{4n^2+1}{8n^3}$$

C
$$\dfrac{4n^2+1}{n^3}$$

D
$$\dfrac{2n^2+1}{2n^3}$$

Solution

$${ T }_{ 2n }=\cfrac { { (2n) }^{ 2 }+1 }{ { (2n) }^{ 3 } } =\cfrac { 4{ n }^{ 2 }+1 }{ 8{ n }^{ 3 } } $$
Question 8
1 / -0

The value of $$\dfrac {1}{i} + \dfrac {1}{i^{2}} + \dfrac {1}{i^{3}} + .... + \dfrac {1}{i^{102}}$$ is

A
$$-1 - i$$

B
$$-1 + i$$

C
$$1 - i$$

D
$$1 + i$$

E
$$1 - 2i$$

Solution

$$\dfrac {1}{i} + \dfrac {1}{i^{2}} + \dfrac {1}{i^{3}} + .... + \dfrac {1}{i^{102}}$$ (GP)
Therefore, $$ S_{n} = \dfrac {\dfrac {1}{i}\left \{1 - \left (\dfrac {1}{i}\right )^{102}\right \}}{1 - \left (\dfrac {1}{i}\right )} = \dfrac {(1 + 1)}{i - 1} \left (\because i^{2} = 1\ and\ i^{2} = -1\right )$$
$$= \dfrac {2(i + 1)}{i^{2} - 1} = -(i + 1) = -1 - i$$
Question 9
1 / -0

The sum of the series $$\displaystyle \sum_{n = 8}^{17} \dfrac{1}{(n + 2)(n + 3)} $$ is equal to

A
$$\dfrac{1}{17}$$

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

C
$$\dfrac{1}{19}$$

D
$$\dfrac{1}{20}$$

E
$$\dfrac{1}{21}$$

Solution

$$\sum_{n=8}^{17}\cfrac1{(n+2)(n+3)}$$
$$=\sum_{n=8}^{17}\cfrac1{(n+2)}-$$$$\cfrac1{(n+3)}$$
$$=\sum_{n=8}^{17}\cfrac1{(n+2)}-$$$$\sum_{n=8}^{17}\cfrac1{(n+3)}$$
$$=\left(\cfrac1{10}+\cfrac1{11}+.....+\cfrac1{19}\right)-$$$$\left(\cfrac1{11}+\cfrac1{12}+.....+\cfrac1{20}\right)$$
$$=\cfrac1{10}-\cfrac1{20}$$
$$=\cfrac1{20}$$
Hence, D is the correct option.
Question 10
1 / -0

Let $$a_n=\dfrac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}$$ then $$\displaystyle \sum_{n=1}^{144}a_n$$ equals

A
$$2456$$

B
$$2645$$

C
$$2466$$

D
$$2546$$

Solution

Given $${ a }_{ n }=\cfrac { 4n+\sqrt { 4{ n }^{ 2 }-1 } }{ \sqrt { 2n+1 } +\sqrt { 2n-1 } } $$
Multiplying Dividing by $$\sqrt { 2n+1 } -\sqrt { 2n-1 } $$

$${ a }_{ n }=\cfrac { \left( 4n+\sqrt { 4{ n }^{ 2 }-1 } \right) \left( \sqrt { 2n+1 } -\sqrt { 2n-1 } \right) }{ \left( 2n+1 \right) -\left( 2n-1 \right) } $$
$${ a }_{ n }=\cfrac { 4n\sqrt { 2n+1 } -4n\sqrt { 2n-1 } +(2n+1)\sqrt { 2n-1 } -(2n-1)\sqrt { 2n+1 } }{ 2 } $$
$$a_n=\cfrac{(2n+1)\sqrt { 2n+1 } -(2n-1)\sqrt { 2n-1 } }{ 2 }$$
$${ a }_{ n }=\cfrac { { \left( 2n+1 \right) }^{ { 3 }/{ 2 } }-{ \left( 2n-1 \right) }^{ { 3 }/{ 2 } } }{ 2 } $$

$$a_1=\cfrac { { \left( 3 \right) }^{ { 3 }/{ 2 } }-{ \left( 1 \right) }^{ { 3 }/{ 2 } } }{ 2 }$$
$$a_2=\cfrac { { \left( 5 \right) }^{ { 3 }/{ 2 } }-{ \left( 3 \right) }^{ { 3 }/{ 2 } } }{ 2 }$$
$$a_3=\cfrac { { \left( 7\right) }^{ { 3 }/{ 2 } }-{ \left( 5 \right) }^{ { 3 }/{ 2 } } }{ 2 }$$
$$a_n=\cfrac { { \left( 2n+1 \right) }^{ { 3 }/{ 2 } }-{ \left( 2n-1 \right) }^{ { 3 }/{ 2 } } }{ 2 }$$

$$\sum _{ i=1 }^{n }{ { a }_{ n } }=\cfrac { { \left( 2n+1 \right) }^{ { 3 }/{ 2 } }-1 }{ 2 }$$
$$\sum _{ i=1 }^{144 }{ { a }_{ n } }$$$$=\cfrac { { \left( 2\times 144+1 \right) }^{ { 3 }/{ 2 } }-1 }{ 2 }=\cfrac { { \left( 289 \right) }^{ { 3 }/{ 2 } }-1 }{ 2 } \\ =\cfrac { { \left( 17\right) }^{ { 3 } }-1 }{ 2 }=2456$$

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Re-Attempt Weekly Quiz Competition

Sequences and Series Test - 28

Result

Sequences and Series Test - 28

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SHARING IS CARING

Question 1

$$e$$

$$5e$$

$$\cfrac{5e}{6} -\cfrac{1}{2}$$

$$\cfrac{5e}{6}$$

Solution

Question 2

$$S< 8$$

$$S> 12$$

$$8< S< 12$$

$$S=8$$

Solution

Question 3

$${e}^{1/2}$$

$$e^2$$

e

$$\frac{1}{e}$$

Solution

Question 4

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

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

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

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

Solution

Question 5

$$9000 < N < 10000$$

$$8000 < N < 9000$$

$$7000 < N < 8000$$

$$10000 < N < 12000$$

Solution

Question 6

Figure 32

Figure 95

Figure 49

Figure 48

Solution

Question 7

$$\dfrac{n^2+1}{8n^3}$$

$$\dfrac{4n^2+1}{8n^3}$$

$$\dfrac{4n^2+1}{n^3}$$

$$\dfrac{2n^2+1}{2n^3}$$

Solution

Question 8

$$-1 - i$$

$$-1 + i$$

$$1 - i$$

$$1 + i$$

$$1 - 2i$$

Solution

Question 9

$$\dfrac{1}{17}$$

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

$$\dfrac{1}{19}$$

$$\dfrac{1}{20}$$

$$\dfrac{1}{21}$$

Solution

Question 10

$$2456$$

$$2645$$

$$2466$$

$$2546$$

Solution

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