Sequences and Series Test

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Sequences and Series Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

For some natural $$N$$ , the number of positive integral $$x$$ satisfying the equation,
$$1!+2!+3!+......+(x)!=(N)^2$$ is :

A
None

B
One

C
Two

D
Infinite

Solution

$$1!+2!+3!+......+(x)!=(N)^2$$ is true for $$x=1, x=3$$ in that case $$N=1,N=3$$ respectively.
For, $$x=4$$, we have $$1!+2!+3!+4!=33$$ which is not a perfect square.
Again for $$x\geq 5$$ $$1!+2!+....+(x)!$$ is of the form $$10k+3 $$ for $$k$$ is some natural number. In these cases, the given sum is not going to be a perfect square.
So, two values of $$x$$ satisfies the given equation.
Question 2
1 / -0

Arrange these numbers in ascending order.
$$756, 567, 657, 676$$

A
$$657, 567, 756, 676$$

B
$$567, 657, 676, 756$$

C
$$756, 676, 657, 567$$

D
$$676, 756, 567, 657$$

Solution

$$\Rightarrow$$ Numbers are said to be in ascending order when they are arranged from the smallest to the largest number.
$$\Rightarrow$$ The numbers which we have to arrange in ascending order are $$756,\,567,\,657$$ and $$676$$
$$\Rightarrow$$ $$567<657<676<756$$
$$\therefore$$ Ascending order $$=567,\,657,\,676,\,756$$
Question 3
1 / -0

$$If\quad { sin }^{ -1 }\left( x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 4 } -......\infty \right) +{ cos }^{ -1 }\left( { x }^{ 2 }-\frac { { x }^{ 4 } }{ 2 } +\frac { { x }^{ 6 } }{ 4 } -.....\infty \right) =\frac { \pi }{ 2 } \quad and\quad 0<x<\sqrt { 2 } \quad then\quad x=$$

A
$$\frac { 1 }{ 2 } $$

B
$$1$$

C
$$-\frac { 1 }{ 2 } $$

D
$$-1$$

Solution

$$x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 4 } +...\infty ={ x }^{ 2 }-\frac { { x }^{ 4 } }{ 2 } +\frac { { x }^{ 6 } }{ 4 } -...\infty \quad x={ x }^{ 2 }\Rightarrow x=1$$
Question 4
1 / -0

If $$S_n=\sum\limits_{r=1}^n \dfrac{2r+1}{r^4+2r^3+r^2}$$ then $$S_{20}$$ =

A
$$\dfrac{220}{221}$$

B
$$\dfrac{420}{441}$$

C
$$\dfrac{439}{221}$$

D
$$\dfrac{440}{441}$$

Solution

$$\dfrac{2r+1}{r^4+2r^3+r^2}=\dfrac{2r+1}{r^2(r+1)^2}=\dfrac{(r+1)^2-r^2}{r^2(r+1)^2}$$
=$$\dfrac{1}{r^2}-\dfrac{1}{(r+1)^2} \therefore \sum\limits_{r=1}^{20}\Big[\dfrac{1}{r^2}-\dfrac{1}{(r+1)^2}\Big]$$
Put r=1, 2, 3, .........20 and terms will cancel diagonally.
$$\therefore \sum \dfrac{1}{1}-\dfrac{1}{(20+1)^2}=1-\dfrac{1}{441}=\dfrac{440}{441}$$.
Question 5
1 / -0

The sum to $$50$$ terms of the series $$\dfrac {1}{2} + \dfrac {3}{4} + \dfrac {7}{8} + \dfrac {15}{16} + ....$$ is equal to

A
$$2^{50} - 51$$

B
$$1 - 2^{-50}$$

C
$$2^{-50}+49$$

D
$$2^{50} - 1$$

Solution

$$T_{n} = \dfrac {2^{n} - 1}{2^{n}} = 1 - \left (\dfrac {1}{2}\right )^{n}$$
$$\therefore S_{n} = n - \displaystyle \sum_{n = 1}^{n} \left (\dfrac {1}{2}\right )^{n} = n - \dfrac {\dfrac {1}{2} . [1 - (1/2)^{n}]}{1 - 1/2}$$
$$= n + 2^{-n} - 1$$.
Question 6
1 / -0

$$\sum\limits_{r=1}^{50}\Big[ \dfrac{1}{49+r}-\dfrac{1}{2r(2r-1)}\Big]=$$

A
$$\dfrac{1}{50}$$

B
$$\dfrac{1}{99}$$

C
$$\dfrac{1}{100}$$

D
$$\dfrac{1}{101}$$

Solution

$$\sum\limits_{r=1}^{50} \cfrac{1}{2r(2r-1)}=\sum\limits_{r=1}^{50}\Big(\cfrac{1}{2r-1}-\cfrac{1}{2r}\Big)$$
$$=\Big( 1-\cfrac{1}{2}+\cfrac{1}{3}-\cfrac{1}{4}+........+\cfrac{1}{99}-\cfrac{1}{100}\Big)$$
$$= \Big(1+\cfrac{1}{2}+\cfrac{1}{3}+\cfrac{1}{4}+.........+\cfrac{1}{100}\Big)-2\Big(\cfrac{1}{2}+\cfrac{1}{4}+............+\cfrac{1}{100}\Big)$$
$$=\sum\limits{r=1}^{100}\cfrac{1}{r}-\Big(1+\cfrac{1}{2}+.....+\cfrac{1}{50}\Big)$$, i.e.,$$\sum\limits_{r=1}^{50}\cfrac{1}{r}=\sum\limits_{r=51}^{100}\cfrac{1}{r}$$
$$\therefore E=\sum\limits_{r=1}^{50}\cfrac{1}{49+r}-\sum\limits_{r=51}^{100}\cfrac{1}{r}$$
$$=\Big[\cfrac{1}{50}+\cfrac{1}{51}+...........+\cfrac{1}{99}\Big] -[\cfrac{1}{51}+\cfrac{1}{52}+............+\cfrac{1}{100}\Big]$$
$$=\cfrac{1}{50}-\cfrac{1}{100}=\cfrac{2}{100}-\cfrac{1}{100}=\cfrac{1}{100}\Rightarrow (C)$$
Question 7
1 / -0

The value of $$x$$ satisfying the equation $$\dfrac{5050-\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+......+\dfrac{5049}{5050}\right)}{1+\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{5050}}=\dfrac{x}{5050}$$ is

A
$$1$$

B
$$5049$$

C
$$5050$$

D
$$5051$$

Solution

First we solve,
$$\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.......+\dfrac{5049}{5050}\right)$$
We can conclude general term or $$n^{th}$$ term,
$$=\sum_{n=1}^{5049}\dfrac{n}{n+1}$$
$$=\sum_{n=1}^{5049}\dfrac{n+1-1}{n+1}$$
$$=\sum_{n=1}^{5049}\left ( 1-\dfrac{1}{n+1} \right )$$

$$=(1-\dfrac{1}{1+1})+(1-\dfrac{1}{2+1})+.......+(1-\dfrac{1}{5049+1})$$
$$=(1-\dfrac{1}{2})+(1-\dfrac{1}{3})+.......+(1-\dfrac{1}{5050})$$
$$=5049-(\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{5050})$$
Substituting this value in $$LHS$$ of the given expression,
$$\dfrac{5050-(\dfrac{1}{2}+\dfrac{2}{3}+.......+\dfrac{5049}{5050})}{1+\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{5050}}$$
$$=\dfrac{5050-5049+(\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{5050})}{1+\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{5050}}$$
$$=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{5050}}{1+\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{5050}}$$
$$=1$$
Now,
$$\dfrac{x}{5050}=1$$

$$x=5050$$
This is the required value of $$x$$.
Question 8
1 / -0

If $$\left| a \right| < 1$$ and $$\left| b \right| < 1$$ then $$\eqalign{ & \cr & S = 1 + \left( {1 + a} \right)b + \left( {1 + a + {a^2} } \right){b^2} + ...} $$=

A
$${1 \over {\left( {1 - b} \right)\left( {1 - ab} \right)}}$$

B
$${1 \over {\left( {1 + b} \right)\left( {1 - ab} \right)}}$$

C
$${1 \over {\left( {1 - b} \right)\left( {1 + ab} \right)}}$$

D
$${1 \over {\left( {1 + b} \right)\left( {1 + ab} \right)}}$$

Solution

Let $$S = 1 + (a + a)b + (1-a+a^2)b^2 + (a+a+a^2+a^3)b^3 + ....$$

then $$bs = b + (1+a)b^2 + (1+a+a^2)b^3 + (1+a+a^2 + a^3)b^4+.......$$

Apply $$(S-bS)$$:

$$S-bS = 1 + b+ab + b^2 + ab^2+a^2-b^2+b^3+ab^3+a^2b^3 + a^3b^2 + .....$$

$$-b-b^2 - ab^2 - b^3 - ab^3-a^2b^3-b^4-ab^4 - a^2b^4-a^3b^4$$ ......

$$S-bS = 1 + ab + a^2b^2 + a^3b^3 +.....$$

the above series is in G.P.

$$S(1-b) = \dfrac{1}{(1-ab)}$$

$$S = \dfrac{1}{(1-ab)(1-b)}$$

$$\therefore$$ option A is correct.
Question 9
1 / -0

If $$\left| a \right| < 1$$ and $$\left| b \right| < 1$$ then $$s = 1 + \left( {1 + a} \right)b + \left( {1 + a + {a^2}} \right){b^2} + \left( {1 + a + {a^2} + {a^3}} \right){b^3} + ...is - .$$

A
$${\dfrac 1 {\left( {1 - b} \right)\left( {1 - ab} \right)}}$$

B
$${\dfrac 1 {\left( {1 + b} \right)\left( {1 - ab} \right)}}$$

C
$${\dfrac 1 {\left( {1 - b} \right)\left( {1 + ab} \right)}}$$

D
$${\dfrac 1 {\left( {1 + b} \right)\left( {1 + ab} \right)}}$$

Solution

$$S=(1+b+b^2+b^3+....)+ab(1+b+b^2+....)+a^2b^2(1+b+b^2+...)+...$$
Since using sum to infinity for G.P we get
$$S=\cfrac{1}{1-b}+\cfrac{ab}{1-b}+\cfrac{a^2b^2}{1-b}+....\\S=\cfrac{1}{1-b}(1+ab+a^2b^2+....)$$
Using G.P for sum to infinity, we get
$$S=\cfrac{1}{1-b}\times\cfrac{1}{1-ab}=\cfrac{1}{(1-b)(1-ab)}$$
Question 10
1 / -0

If $$1 + {x^2} = \sqrt {3}x $$, then $$\displaystyle \prod\limits_{n = 1}^{24} {\left( {x^n} + \dfrac {1} {x^n} \right)} $$ is equal to

A
$$48$$

B
$$-48$$

C
$$ \pm 48\left( {\omega - {\omega ^2}} \right)$$

D
none of these

Solution

$$1+{x}^{2}=\sqrt{3}x$$
$$\Rightarrow\,\dfrac{1}{x}+x=\sqrt{3}$$
$$\Rightarrow\,x+\dfrac{1}{x}=\sqrt{3}$$
$$\Rightarrow\,{\left(x+\dfrac{1}{x}\right)}^{2}={\left(\sqrt{3}\right)}^{2}$$
$$\Rightarrow\,{x}^{2}+\dfrac{1}{{x}^{2}}+2x\times \dfrac{1}{x}=3$$
$$\Rightarrow\,{x}^{2}+\dfrac{1}{{x}^{2}}+2=3$$
$$\Rightarrow\,{x}^{2}+\dfrac{1}{{x}^{2}}=3-2=1$$
$$\Rightarrow\,{\left(x+\dfrac{1}{x}\right)}^{3}={\left(\sqrt{3}\right)}^{3}$$
$$\Rightarrow\,{x}^{3}+\dfrac{1}{{x}^{3}}+3x\times \dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=3\sqrt{3}$$
$$\Rightarrow\,{x}^{3}+\dfrac{1}{{x}^{3}}+3\left(\sqrt{3}\right)=3\sqrt{3}$$ since $$x+\dfrac{1}{x}=\sqrt{3}$$
$$\Rightarrow\,{x}^{3}+\dfrac{1}{{x}^{3}}=3\sqrt{3}-3\sqrt{3}=0$$
$$\displaystyle\prod_{n=1}^{24}{{x}^{n}+\dfrac{1}{{x}^{n}}}=\left(x+\dfrac{1}{x}\right)\left({x}^{2}+\dfrac{1}{{x}^{2}}\right)\left({x}^{3}+\dfrac{1}{{x}^{3}}\right)...\left({x}^{24}+\dfrac{1}{{x}^{24}}\right)=0$$ since $${x}^{3}+\dfrac{1}{{x}^{3}}=0$$

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Re-Attempt Weekly Quiz Competition

Sequences and Series Test - 35

Result

Sequences and Series Test - 35

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SHARING IS CARING

Question 1

None

One

Two

Infinite

Solution

Question 2

$$657, 567, 756, 676$$

$$567, 657, 676, 756$$

$$756, 676, 657, 567$$

$$676, 756, 567, 657$$

Solution

Question 3

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

$$1$$

$$-\frac { 1 }{ 2 } $$

$$-1$$

Solution

Question 4

$$\dfrac{220}{221}$$

$$\dfrac{420}{441}$$

$$\dfrac{439}{221}$$

$$\dfrac{440}{441}$$

Solution

Question 5

$$2^{50} - 51$$

$$1 - 2^{-50}$$

$$2^{-50}+49$$

$$2^{50} - 1$$

Solution

Question 6

$$\dfrac{1}{50}$$

$$\dfrac{1}{99}$$

$$\dfrac{1}{100}$$

$$\dfrac{1}{101}$$

Solution

Question 7

$$1$$

$$5049$$

$$5050$$

$$5051$$

Solution

Question 8

$${1 \over {\left( {1 - b} \right)\left( {1 - ab} \right)}}$$

$${1 \over {\left( {1 + b} \right)\left( {1 - ab} \right)}}$$

$${1 \over {\left( {1 - b} \right)\left( {1 + ab} \right)}}$$

$${1 \over {\left( {1 + b} \right)\left( {1 + ab} \right)}}$$

Solution

Question 9

$${\dfrac 1 {\left( {1 - b} \right)\left( {1 - ab} \right)}}$$

$${\dfrac 1 {\left( {1 + b} \right)\left( {1 - ab} \right)}}$$

$${\dfrac 1 {\left( {1 - b} \right)\left( {1 + ab} \right)}}$$

$${\dfrac 1 {\left( {1 + b} \right)\left( {1 + ab} \right)}}$$

Solution

Question 10

$$48$$

$$-48$$

$$ \pm 48\left( {\omega - {\omega ^2}} \right)$$

none of these

Solution

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