Sequences and Series Test

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Sequences and Series Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

The $$(n+1)^{th} $$ term from the end in $$(x - \frac{1}{x})^{3n}$$ is

A
$$3nc_n.X^{-n}$$

B
$$(-1)^n. 3nc_n .X^{-n}$$

C
$$3nc_nX^n$$

D
$$(-1)^n. 3nc_n.X^n$$
Question 2
1 / -0

The sume of the series $$1^3 - 2^3 + 3^3 - ........ + 9^3$$ =

A
300

B
125

C
425

D
0

Solution
Question 3
1 / -0

If $$a_{1}, a_{2}, ......., a_n(n > 3)$$ are all unequal positive real numbers, and

$$E = \dfrac{(1 + a_{1} + a_{1}^{2})(1 + a_{2} + a_{2}^{2})......(1 + a_{n} + a_{n}^{2})}{a_{1}, a_{2}, ......., a_{n}}$$
then which of the following best describes E?

A
$$E \leq 2^{n}$$

B
$$E \geq 3^{n}$$

C
$$E > 3^{n}$$

D
$$E > 2^{n}$$

Solution

$${ a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },...{ a }_{ n }$$ all are unequal $$n>3\\ E=\cfrac { \left( 1+{ a }_{ 1 }+{ a }_{ 1 }^{ 2 } \right) \left( 1+{ a }_{ 2 }+{ a }_{ 2 }^{ 2 } \right) ...\left( 1+{ a }_{ n }+{ a }_{ n }^{ n } \right) }{ { a }_{ 1 }.{ a }_{ 2 }.{ a }_{ 3 }....{ a }_{ n } } \\ =\left( \cfrac { 1 }{ { a }_{ 1 } } +1+{ a }_{ 1 } \right) \left( 1+\cfrac { 1 }{ { a }_{ 2 } } +{ a }_{ 2 } \right) \left( 1+\cfrac { 1 }{ { a }_{ 3 } } +{ a }_{ 3 } \right) ....\left( 1+\cfrac { 1 }{ { a }_{ n } } +{ a }_{ n } \right) \\ A.M.\ge G.M.\\ \cfrac { \cfrac { 1 }{ { a }_{ 1 } } +{ a }_{ 1 }+1 }{ 3 } \ge \sqrt [ 3 ]{ \cfrac { 1 }{ { a }_{ 1 } } .{ a }_{ 1 }.1 } \\ \cfrac { 1 }{ { a }_{ 1 } } +{ a }_{ 1 }+1\ge 3$$
As $${ a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },...{ a }_{ n }$$ are unequal if $$\cfrac { 1 }{ { a }_{ 1 } } +{ a }_{ 1 }+1=3$$,then all others must be $$>3\\ E=\left( \cfrac { 1 }{ { a }_{ 1 } } +1+{ a }_{ 1 } \right) \left( 1+\cfrac { 1 }{ { a }_{ 2 } } +{ a }_{ 2 } \right) ....\left( 1+\cfrac { 1 }{ { a }_{ n } } +{ a }_{ n } \right) \\ >3.3.....3(ntimes)\\ E>{ 3 }^{ n }$$
Question 4
1 / -0

The sum of the series $${1 \over 2} + {3 \over 4} + {7 \over 8} + {{15} \over {16}} + .......$$ up to low upon to n term is to the

A
$$n - 1 + {2^{ - n}}$$

B
$${2^{ - n}} - 1$$

C
$$n - 2$$

D
None of these

Solution

$$\cfrac { 1 }{ 2 } +\cfrac { 3 }{ 4 } +\cfrac { 7 }{ 8 } +\cfrac { 15 }{ 16 } +....upto\quad n\quad terms\\ =(1-\cfrac { 1 }{ 2 } )+(1-\cfrac { 1 }{ 4 } )+....+(1-\cfrac { 1 }{ { 2 }^{ n } } )\\ =n-(\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } +....+\cfrac { 1 }{ { 2 }^{ n } } )\\ =n-\cfrac { \cfrac { 1 }{ 2 } \times \left( { \left( \cfrac { 1 }{ 2 } \right) }^{ n }-1 \right) }{ \cfrac { 1 }{ 2 } -1 } \\ =n+{ 2 }^{ -n }-1$$
Question 5
1 / -0

The positive integer n for which $$2 \times {2^2} + 3 \times {2^3} + 4 \times {2^4} + ....... + n \times {2^n} = {2^{^{n + 10}}}$$ is______

A
$$510$$

B
$$511$$

C
$$512$$

D
$$513$$

Solution

$$2 \times 2^2 + 3 \times 2^3 + 4 \times 2^4 ... n \times 2^n = 2^{n + 10}$$
It is $$AGP$$ since
$$2, 3, 4 ... AP$$
$$2^3 , 2^3, 2^4 .... GP$$
let
$$S = 2 \times 2^2 + 3.2^3 + 4.2^4 ... n.2^n$$
$$2S = $$ $$2.2^3 + 3.2^4 ... (n - 1). 2^n + n . 2^{n + 1}$$
__________________________________________
$$-S = 2.2^2 + 2^3 + 2^4 ... + 2^n - n . 2^{n + 1}$$
$$-S = 2.2^2 + 2^3 \left(\dfrac{2^{n-3} - 1}{2-1} \right) - n. 2^{n + 1}$$
$$-S = 2.2^2 + 2^3 (2^{n - 3} - 1) - n.2^{n +1}$$
$$= 2^3 + 2^n - 2^3 - n.2^{n+1}$$
$$= 2^n - 2n.2^n$$
$$S = 2^n (2n - 1)$$
$$2^n . 2^{10} = 2^n (2n - 1)$$
$$2^{10} = 2n - 1$$
$$1024 = 2n - 1$$
$$1025 = 2n$$
$$n = \dfrac{1025}{2}$$
$$n = 512.5$$
integer $$= 512$$
Question 6
1 / -0

The sum of infinite series $$\dfrac{1.3}{2}+\dfrac{3.5}{2^2}+\dfrac{5.7}{2^3}+\dfrac{7.9}{2^4}+...\infty$$.

A
$$21$$

B
$$22$$

C
$$23$$

D
None

Solution

$$S=\cfrac { 1.3 }{ 2 } +\cfrac { 3.5 }{ { 2 }^{ 2 } } +\cfrac { 5.7 }{ { 2 }^{ 3 } } +\cfrac { 7.9 }{ { 2 }^{ 4 } } +....\infty \\ \cfrac { S }{ 2 } =\cfrac { 1.3 }{ { 2 }^{ 2 } } +\cfrac { 3.5 }{ { 2 }^{ 3 } } +\cfrac { 5.7 }{ { 2 }^{ 4 } } +....\infty \\ S-\cfrac { S }{ 2 } =\cfrac { 3 }{ 2 } +\cfrac { 12 }{ { 2 }^{ 2 } } +\cfrac { 20 }{ { 2 }^{ 3 } } +\cfrac { 28 }{ { 2 }^{ 4 } } +....\infty \\ \cfrac { S }{ 2 } =\cfrac { 3 }{ 2 } +4\left( \cfrac { 3 }{ { 2 }^{ 2 } } +\cfrac { 5 }{ { 2 }^{ 3 } } +\cfrac { 7 }{ { 2 }^{ 4 } } +....\infty \right) \\ \cfrac { S }{ 2 } =\cfrac { 3 }{ 2 } +3+\cfrac { 5 }{ { 2 } } +\cfrac { 7 }{ { 2 }^{ 2 } } +\cfrac { 9 }{ { 2 }^{ 3 } } +....\infty \\ \cfrac { S }{ 4 } =\cfrac { 3 }{ 4 } +\cfrac { 3 }{ 2 } +\cfrac { 5 }{ { 2 }^{ 2 } } +\cfrac { 7 }{ { 2 }^{ 3 } } +....\infty \\ \cfrac { S }{ 2 } -\cfrac { S }{ 4 } =\cfrac { 15 }{ 4 } +\left( \cfrac { 2 }{ 2 } +\cfrac { 2 }{ { 2 }^{ 2 } } +\cfrac { 2 }{ { 2 }^{ 3 } } +....\infty \right) \\ \cfrac { S }{ 4 } =\cfrac { 15 }{ 4 } +\cfrac { 1 }{ 1-\cfrac { 1 }{ 2 } } \\ S=23$$
Question 7
1 / -0

The sum to infinite of the series
$$1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + ........$$

A
2

B
3

C
4

D
6

Solution

$$s=1+\cfrac { 2 }{ 3 } +\cfrac { 6 }{ 3^{ 2 } } +\cfrac { 10 }{ 3^{ 3 } } +....\\ s-1=\cfrac { 2 }{ 3 } +\cfrac { 6 }{ 3^{ 2 } } +\cfrac { 10 }{ 3^{ 3 } } +...\\ \cfrac { s-1 }{ 3 } =\cfrac { 2 }{ 3^{ 2 } } +\cfrac { 6 }{ 3^{ 3 } } +\cfrac { 10 }{ 3^{ 4 } } +...\\ (s-1)-(\cfrac { s-1 }{ 3 } )=\cfrac { 2 }{ 3 } +\cfrac { 4 }{ 3^{ 2 } } +\cfrac { 4 }{ 3^{ 3 } } +\cfrac { 4 }{ 3^{ 4 } } +......\\ (s-1)-(\cfrac { s-1 }{ 3 } )=\cfrac { 2 }{ 3 } +\cfrac { 4 }{ 3^{ 2 } } [1+\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 3^{ 2 } } +\cfrac { 1 }{ 3^{ 3 } } +....]\\ \cfrac { 3s-3-s+1 }{ 3 } =\cfrac { 2 }{ 3 } +\cfrac { 4 }{ 3^{ 2 } } (\cfrac { 1 }{ 1 } -\cfrac { 1 }{ 3 } )\\ \cfrac { 2 }{ 3 } (s-1)=2\times \cfrac { 2 }{ 3 } \\ s-1=2\\ s=3$$
Question 8
1 / -0

If $$ \displaystyle \lim _{ x\rightarrow 0^+ }{ x\left( \left[ \dfrac { 1 }{ x } \right] +\left[ \dfrac { 5 }{ x } \right] +\left[ \dfrac { 11 }{ x } \right] +\left[ \dfrac { 19 }{ x } \right] +\left[ \dfrac { 29 }{ x } \right] +.......to\quad n\quad terms \right) }=430$$ (where [.] denotes the greatest integer function), then $$n=$$

A
$$8$$

B
$$9$$

C
$$10$$

D
$$11$$

Solution

$$\lim_{x\rightarrow 0^{+}} x\left( \left[\dfrac{1}{x}\right] + \left[\dfrac{5}{x}\right] + \left[\dfrac{11}{x}\right] + \left[\dfrac{19}{x}\right] + \left[\dfrac{29}{x}\right] + ....... to \quad n \quad terms\right) = 430$$

$$\implies 1 + 5 + 11 + 19 + 29 + ......... n terms = 430$$

$$\implies 1 + 5 + 11 + 19 + 29 + .........+ t_n = 430$$

$$1 + 5 + 11 + 19 + 29 + ......... t_n$$
By method of difference,
$$t_n = n^2 + n - 1$$
$$\because S_n = \sum t_n$$

$$\implies \sum t_n = 430$$

$$\implies \sum (n^2 + n - 1) = 430$$

$$\implies \sum n^2 + \sum n - \sum 1 = 430$$

$$\implies \dfrac{n(n+1)(2n+1)}{6} + \dfrac{n(n+1)}{2} - n = 430$$

$$\implies \dfrac{n(n+1)}{2} \left[ \dfrac{2n+1}{3} + 1\right] - n = 430$$

$$\implies \dfrac{n(n+1)}{2} \left[ \dfrac{2n+4}{3} \right] - n = 430$$

$$\implies \dfrac{n(n+1)(n+2)}{3} - n = 430$$

$$\implies n(n+1)(n+2) - 3n = 430\times 3$$

$$\implies n^3 + 3n^2 + 2n - 3n = 1290$$

$$\implies n (n^2 + 3n - 1 ) = 10\times 129$$

$$\implies n (n^2 + 3n - 1 ) = 10\times (10^2 + 30 - 1)$$

$$\therefore n = 10$$

Option C is correct.
Question 9
1 / -0

If $$\displaystyle\sum _{ n=1 }^{ 2013 }{ \tan { \left( \dfrac { \theta }{ { 2 }^{ n } } \right) } } \sec { \left( \dfrac { \theta }{ { 2 }^{ n-1 } } \right) } =\left( \dfrac { \theta }{ { 2 }^{ a } } \right) -\left( \dfrac { \theta }{ { 2 }^{ b } } \right)$$ then $$(b+a)$$ equals

A
$$2014$$

B
$$2012$$

C
$$2013$$

D
$$2019$$

Solution
Question 10
1 / -0

If $$\sum^5_{n=1}\dfrac{1}{n(n+1)(n+2)(n+3)}=\dfrac{k}{3}$$, then k is equal to?

A
$$\dfrac{55}{336}$$

B
$$\dfrac{17}{105}$$

C
$$\dfrac{19}{112}$$

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

Solution

$$\\ \sum _{ n=1 }^{ 5 }{ \cfrac { 1 }{ n(n+1)(n+2)(n+3) } =\cfrac { k }{ 3 } } \\ \cfrac { 1 }{ 2 } \sum _{ n=1 }^{ 5 }{ \left( \cfrac { 1 }{ n(n+3) } -\cfrac { 1 }{ (n+1)(n+3) } \right) } =\cfrac { k }{ 3 } \\ \\ \sum _{ n=1 }^{ 5 }{ \left[ \cfrac { 1 }{ 3 } \left( \cfrac { 3 }{ n(n+3) } \right) -\cfrac { 1 }{ (n+1)(n+3) } \right] } =\cfrac { 2k }{ 3 } \\ \sum _{ n=1 }^{ 5 }{ \left( \cfrac { 1 }{ 3n } -\cfrac { 1 }{ n+1 } +\cfrac { 1 }{ n+2 } -\cfrac { 1 }{ 3(n+3) } \right) } =\cfrac { 2k }{ 3 } \\ \cfrac { 1 }{ 3 } \left( 1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 5 } \right) -\left( \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 5 } +\cfrac { 1 }{ 6 } \right) +\left( \cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 5 } +\cfrac { 1 }{ 6 } +\cfrac { 1 }{ 7 } \right) -\cfrac { 1 }{ 3 } \left( \cfrac { 1 }{ 4 } +\cfrac { 1 }{ 5 } +\cfrac { 1 }{ 6 } +\cfrac { 1 }{ 7 } +\cfrac { 1 }{ 8 } \right) =\cfrac { 2k }{ 3 } \\ \cfrac { 1 }{ 3 } \left( 1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } -\cfrac { 1 }{ 6 } -\cfrac { 1 }{ 7 } -\cfrac { 1 }{ 8 } \right) -\left( \cfrac { 1 }{ 2 } -\cfrac { 1 }{ 7 } \right) =\cfrac { 2k }{ 3 } \\ \cfrac { 1 }{ 3 } \left( \cfrac { 235 }{ 3\times 56 } \right) -\cfrac { 5 }{ 14 } =\cfrac { 2k }{ 3 } \\ \cfrac { 235 }{ 3\times 56 } -\cfrac { 5 }{ 14 } =2k\\ \cfrac { 55 }{ 3\times 56 } =2k\\ k=\cfrac { 55 }{ 336 } $$

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Re-Attempt Weekly Quiz Competition

Sequences and Series Test - 56

Result

Sequences and Series Test - 56

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SHARING IS CARING

Question 1

$$3nc_n.X^{-n}$$

$$(-1)^n. 3nc_n .X^{-n}$$

$$3nc_nX^n$$

$$(-1)^n. 3nc_n.X^n$$

Question 2

300

125

425

0

Solution

Question 3

$$E \leq 2^{n}$$

$$E \geq 3^{n}$$

$$E > 3^{n}$$

$$E > 2^{n}$$

Solution

Question 4

$$n - 1 + {2^{ - n}}$$

$${2^{ - n}} - 1$$

$$n - 2$$

None of these

Solution

Question 5

$$510$$

$$511$$

$$512$$

$$513$$

Solution

Question 6

$$21$$

$$22$$

$$23$$

None

Solution

Question 7

2

3

4

6

Solution

Question 8

$$8$$

$$9$$

$$10$$

$$11$$

Solution

Question 9

$$2014$$

$$2012$$

$$2013$$

$$2019$$

Solution

Question 10

$$\dfrac{55}{336}$$

$$\dfrac{17}{105}$$

$$\dfrac{19}{112}$$

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

Solution

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