Sequences and Series Test

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Sequences and Series Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

If $$ s_n = \displaystyle \sum_{r=0}^n \dfrac {1}{^nC_r}$$ and $$ t_n = \displaystyle \sum_{r=0}^n \dfrac {r}{^nC_r} , $$ then $$ \dfrac {t_n}{s_n} $$ is equal to -

A
$$ \dfrac {n}{2} $$

B
$$ \dfrac {n}{2} -1 $$

C
$$ n - 1 $$

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

Solution

$${ S }_{ n }={ \sum }_{ r=0 }^{ n }\cfrac { 1 }{ \; ^{ n }{ C }_{ r } } =\cfrac { 1 }{ \; ^{ n }{ C }_{ 0 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { 1 }{ \; ^{ n }{ C }_{ n-1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ n } } \\ { S }_{ n }=1+\left\{ \cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { 1 }{ \; ^{ n }{ C }_{ n-1 } } \right\} +1\\ \; ^{ n }{ C }_{ 0 }=1,\quad \; ^{ n }{ C }_{ n }=1,\quad \; ^{ n }{ C }_{ r }=\; ^{ n }{ C }_{ n-r }\\ { S }_{ n }=2+2\left\{ \cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { 1 }{ \; ^{ n }{ C }_{ n/2 } } \right\} \\ \cfrac { { S }_{ n }-2 }{ 2 } =\left\{ \cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { 1 }{ \; ^{ n }{ C }_{ n/2 } } \right\} \quad \quad \quad (1)\\ { t }_{ n }={ \sum }_{ r=0 }^{ n }\cfrac { r }{ \; ^{ n }{ C }_{ r } } =0+\cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 2 }{ \; ^{ n }{ C }_{ 2 } } +\cfrac { 3 }{ \; ^{ n }{ C }_{ 3 } } +...+\cfrac { n-2 }{ \; ^{ n }{ C }_{ n-2 } } +\cfrac { n-1 }{ \; ^{ n }{ C }_{ n-1 } } +\cfrac { n }{ \; ^{ n }{ C }_{ n } } \\ { t }_{ n }=\cfrac { 1+n-1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 2+n-2 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { n/2+n-n/2 }{ \; ^{ n }{ C }_{ n/2 } } +\cfrac { n }{ 1 } \\ { t }_{ n }=n\left[ \cfrac { 1 }{ \; ^{ n }{ C }_{ 1 } } +\cfrac { 1 }{ \; ^{ n }{ C }_{ 2 } } +...+\cfrac { 1 }{ \; ^{ n }{ C }_{ n/2 } } \right] +n$$

From $$(1)$$
$${ t }_{ n }=n\left( \cfrac { { S }_{ n }-2 }{ 2 } \right) +n\\ { t }_{ n }=n\left( \cfrac { { S }_{ n }-2+2 }{ 2 } \right) =\cfrac { n }{ 2 } { S }_{ n }\\ \cfrac { { t }_{ n } }{ { S }_{ n } } =\cfrac { n }{ 2 } $$
Question 2
1 / -0

$$\displaystyle \sum_{r = 0}^{n}{t^3 \left( \frac{^nC_r}{^nC_{r-1}} \right)^2 }$$ is equal to

A
$$\displaystyle \frac{n(n+1)(n+2)^2}{12} $$

B
$$\displaystyle \frac{n(n+1)^2 (n+2)}{12} $$

C
$$\displaystyle \frac{n(n+1)(n+2)}{12} $$

D
None of these

Solution

We have

$$s=\displaystyle\sum _0^nr^3\Big(\dfrac{^nC_r}{^nC_{r-1}}\Big)^2$$

$$s=\displaystyle\sum _0^nr^3\Big(\dfrac{n-r+1}{r}\Big)^2$$

$$s=\displaystyle\sum _0^nr\Big({n-r+1}\Big)^2$$

$$s=\displaystyle\sum _0^nr(n^2+r^2+1-2nr-2r+2n)$$

$$s=\displaystyle\sum _0^nr(n^2+2n+1)+r^3-2r^2(n+1)$$

$$s=\displaystyle\sum _0^nr(n+1)^2+r^3-2r^2(n+1)$$

$$s=\dfrac{n(n+1)^3}{2}+\dfrac{n^2(n+1)^2}{4}-\dfrac{n(n+1)^2(2n+1)}{3}$$

Taking $$n(n+1)^{2}$$ common, we get

$$s=n(n+1)^{2}\Big(\dfrac{(n+1)}{2}+\dfrac{n}{4}-\dfrac{(2n+1)}{3}\Big)$$

$$s=\dfrac{n(n+1)^2(n+2)}{12}$$

B is the correct answer
Question 3
1 / -0

Solve the given series:
$$\dfrac {1.2^2+2.3^2+3.4^2+...n(n+1)^2}{1.2+2^2.3+3^2.4+...n^2(n+1)}$$

A
$$(3n+1)(3n+5)$$

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

C
$$\dfrac {3n+1}{3n+5}$$

D
$$\dfrac {3n+5}{3n+1}$$

Solution

$$=\dfrac{1.2^{2}+2.3^{2}+3.4^{2}+...+n(n+1)^{2}}{1^{2}.2+2^{2}.3+3^{2}.4+...+n^{2}(n+1)}$$

$$=\dfrac{(2-1).2^{2}+(3-1).3^{2}+(4-1).4^{2}+...+(n+1-1)(n+1)^{2}}{1^{2}.(1+1)+2^{2}.(2+1)+3^{2}.(3+1)+...+n^{2}(n+1)}$$

$$=\dfrac{[1^{3}+2^{3}+3^{3}+4^{3}+...+(n+1)^{3}]-[1^{2}+2^{2}+3^{2}+4^{2}+...+(n+1)^{2}]}{[1^{3}+2^{3}+3^{3}+...+n^{3}]+[1^{2}+2^{2}+3^{2}+...+n^{2}]}$$

$$=\dfrac{\left[\frac{(n+1)^{2}(n+2)^{2}}{4}\right]-\left[\frac{(n+1)(n+2)(2n+3)}{6}\right]}{\left[\frac{(n)^{2}(n+1)^{2}}{4}\right]+\left[\frac{(n)(n+1)(2n+1)}{6}\right]}$$

$$=\dfrac{(n+1)(n+2)[6(n+1)(n+2)-4(2n+3)]}{n(n+1)[6(n)(n+1)+4(2n+1)]}$$

$$=\dfrac{(n+2)[6n^{2}+10n]}{n[6n^{2}+14n+4]}$$

$$=\dfrac{2n(n+2)[3n+5]}{2n[3n^{2}+7n+2]}$$

$$=\dfrac{(n+2)(3n+5)}{(3n+1)(n+2)}$$

$$=\dfrac{(3n+5)}{(3n+1)}$$
Question 4
1 / -0

If $$x=\dfrac{1}{5}+\dfrac{1.3}{5.10}+\dfrac{1.3.5}{5.10.15}+.....\infty$$ then $$3x^2+6x=$$

A
1

B
2

C
3

D
4

Solution
Question 5
1 / -0

The value of $$\sum _{ n=1 }^{ 9999 }{ \cfrac { 1 }{ \left( \sqrt { n } +\sqrt { n+1 } \right) \left( \sqrt [ 4 ]{ n } +\sqrt [ 4 ]{ n+1 } \right) } } $$ is

A
$$9$$

B
$$99$$

C
$$999$$

D
$$9999$$

Solution

Let $${ T }_{ n }=\cfrac { 1 }{ \left( \sqrt { n } +\sqrt { n+1 } \right) \left( \sqrt [ 4 ]{ n } +\sqrt [ 4 ]{ n+1 } \right) } \times \cfrac { \sqrt { n+1 } -\sqrt { n } }{ \sqrt { n+1 } -\sqrt { n } } \\ { T }_{ n }=\cfrac { \sqrt { n+1 } -\sqrt { n } }{ \sqrt [ 4 ]{ n+1 } +\sqrt [ 4 ]{ n } } =\cfrac { \left( \sqrt [ 4 ]{ n+1 } -\sqrt [ 4 ]{ n } \right) \left( \sqrt [ 4 ]{ n+1 } +\sqrt [ 4 ]{ n } \right) }{ \left( \sqrt [ 4 ]{ n+1 } +\sqrt [ 4 ]{ n } \right) } $$
$${ T }_{ n }=\sqrt [ 4 ]{ n+1 } -\sqrt [ 4 ]{ n } \\ { T }_{ 1 }=\sqrt [ 4 ]{ 2 } -\sqrt [ 4 ]{ 1 } \\ { T }_{ 2 }=\sqrt [ 4 ]{ 3 } -\sqrt [ 4 ]{ 2 } \\ { T }_{ 3 }=\sqrt [ 4 ]{ 4 } -\sqrt [ 4 ]{ 3 } $$

$${ T }_{ n }=\sqrt [ 4 ]{ n+1 } -\sqrt [ 4 ]{ n } $$
Adding all
$${ S }_{ n }=\sqrt [ 4 ]{ n+1 } -\sqrt [ 4 ]{ 1 } =\sqrt [ 4 ]{ n+1 } -1$$
$${ S }_{ 9999 }$$$$=\sqrt [ 4 ]{ 9999+1 } -1\\ =\sqrt [ 4 ]{ 10000 } -1\\ =10-1\\ =9$$
Question 6
1 / -0

Let $${ T }_{ r }$$ and $${ S }_{ r }$$ be the $${ r }^{ th }$$ term and sum up to $${ r }^{ th }$$ term of a series respectively. If for an odd natural number $$n,{ S }_{ n }=n$$ and $${ T }_{ n }=\dfrac { { T }_{ n-1 } }{ { n }^{ 2 } }$$, then $${ T }_{ m }$$ ($$m$$ being even) is:

A
$$\dfrac { 2 }{ 1+{ m }^{ 2 } }$$

B
$$\dfrac { 2{ m }^{ 2 } }{ 1+{ m }^{ 2 } }$$

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

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

Solution

Given,
$$S_n=n$$ and $$T_n=\dfrac {T_{n-1}}{n^2},$$ $$n$$ is an odd natural number.

$$\therefore S_n - S_{n-2} =(n)-(n-2)= 2\\ \Rightarrow T_n + T_{n-1} = 2\\ \Rightarrow \dfrac {T_{n-1}}{n^2}+ T_{n-1}= 2\\ \Rightarrow \left( \dfrac{1}{n^2} + 1 \right) T_{n-1} = 2\\ \Rightarrow \left( \dfrac{1+n^2}{n^2} \right) T_{n-1} = 2\\ \Rightarrow T_{n-1} = \dfrac{2n^2}{1 + n^2}$$

Since, $$n$$ is odd $$\Rightarrow (n-1)$$ is odd.

Replacing $$(n-1)$$ by $$m$$ we get,
$$T_m = \dfrac{2(m+1)^2}{1 + (m+1)^2}$$
Question 7
1 / -0

Sum of the series $$\displaystyle\sum^n_{r=1}(r^2+1)r!$$ is?

A
$$(n+1)!$$

B
$$(n+2)!-1$$

C
$$n-(n+1)!$$

D
$$n-(n+2)!$$
Question 8
1 / -0

In a certain code language, DIPLOMA is written as FERHQIC, then what is the code for PENCILS in the language?

A
RAPYHKU

B
RAPYKHU

C
RPAYKHU

D
RAPKYHU
Question 9
1 / -0

If $$x\in R$$ and $$S=1-{ C }_{ 1 }\cfrac { 1+x }{ { \left( 1+nx \right) }^{ } } +{ C }_{ 2 }\cfrac { 1+2x }{ { \left( 1+nx \right) }^{ 2 } } -{ C }_{ 3 }\cfrac { 1+3x }{ { \left( 1+nx \right) }^{ 3 } } +...upto\quad (n+1)$$ terms, then $$S$$

A
equal $${x}^{2}$$

B
equals $$1$$

C
equals $$0$$

D
is independent
Question 10
1 / -0

The sum to infinite of the series
$$S=1+\cfrac { 2 }{ 3 } +\cfrac { 6 }{ { 3 }^{ 2 } } +\cfrac { 6 }{ { 3 }^{ 3 } } +\cfrac { 6 }{ { 3 }^{ 4 } } +.....\quad $$ is

A
$$4$$

B
$$3$$

C
$$2$$

D
$$6$$

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Re-Attempt Weekly Quiz Competition

Sequences and Series Test - 55

Result

Sequences and Series Test - 55

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SHARING IS CARING

Question 1

$$ \dfrac {n}{2} $$

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

$$ n - 1 $$

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

Solution

Question 2

$$\displaystyle \frac{n(n+1)(n+2)^2}{12} $$

$$\displaystyle \frac{n(n+1)^2 (n+2)}{12} $$

$$\displaystyle \frac{n(n+1)(n+2)}{12} $$

None of these

Solution

Question 3

$$(3n+1)(3n+5)$$

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

$$\dfrac {3n+1}{3n+5}$$

$$\dfrac {3n+5}{3n+1}$$

Solution

Question 4

1

2

3

4

Solution

Question 5

$$9$$

$$99$$

$$999$$

$$9999$$

Solution

Question 6

$$\dfrac { 2 }{ 1+{ m }^{ 2 } }$$

$$\dfrac { 2{ m }^{ 2 } }{ 1+{ m }^{ 2 } }$$

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

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

Solution

Question 7

$$(n+1)!$$

$$(n+2)!-1$$

$$n-(n+1)!$$

$$n-(n+2)!$$

Question 8

RAPYHKU

RAPYKHU

RPAYKHU

RAPKYHU

Question 9

equal $${x}^{2}$$

equals $$1$$

equals $$0$$

is independent

Question 10

$$4$$

$$3$$

$$2$$

$$6$$

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