Sequences and Series Test

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Sequences and Series Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

The sum of infinity of the series $$\dfrac{1}{1} + \dfrac{1}{1 + 2} + \dfrac{1}{1+2+3}+$$______ is equal to:

A
$$2$$

B
$$\cfrac{5}{2}$$

C
$$3$$

D
None of these

Solution
Question 2
1 / -0

A series is given as: $$4+7+10+13+16+.....$$ Find the sum of the series up to $$10$$ terms.

A
$$110$$

B
$$125$$

C
$$162$$

D
$$175$$

Solution

The given series is $$4+7+10+13+16+...$$
$$a=4,d=3$$

The sum of series is
$$S_n=\dfrac n2[2a+(n-1)d]$$

$$=\dfrac {10}2[2(4)+9(3)]\\$$
$$=5(8+27)\\$$
$$=5\times 35\\$$
$$=175$$
Question 3
1 / -0

If $$a_n=n(n!)$$, then $$\displaystyle\sum^{100}_{r=1} a_r$$ is equal?

A
$$101!$$

B
$$100!-1$$

C
$$101!-1$$

D
$$101!+1$$

Solution

Given an= $$n(n!)$$
$$=(n+1-1)n!$$
$$=(n+1)n!-n!$$
$$=(n+1)!-n!$$
$$\therefore \displaystyle \sum_{r=1}^{100}a_{r}=a_{1}+a_{2}+a_{3}+.......+a_{100}$$
$$=((1+1)!-1!)+((2+1)!-2!)+....+((100+1)!-100!)$$
$$=2!-1!+3!-2!+.....+100!-99!+101!-100!$$
$$=101!-1$$
Question 4
1 / -0

If $$a_{1}=a_{2}=2,a_{n}=a_{n-1}-1(n > 2)$$ then $$a_{5}$$ is ?

A
$$1$$

B
$$-1$$

C
$$0`$$

D
$$-2$$

Solution

Given that $${ a }_{ 1 }=2\quad and\quad { a }_{ 2 }=2$$

use $${ a }_{ n }={ a }_{ n-1 }-1$$

Putting n=3,$${ a }_{ 3 }={ a }_{ 3-1 }-1$$

$${ a }_{ 3 }={ a }_{ 2 }-1=2-1=1$$

Putting n=4,$${ a }_{ 4 }={ a }_{ 4-1 }-1$$

$${ a }_{ 4 }={ a }_{ 3 }-1=1-1=0$$

Putting n=5,$${ a }_{ 5 }={ a }_{ 5-1 }-1$$

$${ a }_{ 5 }={ a }_{ 4 }-1=0-1=-1$$

Hence the five terms of sequence are $$2,2,1,0 and-1.$$
Question 5
1 / -0

The sum of the series $$1+\dfrac{1}{4\times 2!}+\dfrac{1}{16\times 4!}+\dfrac{1}{64\times 6!}+....\infty$$ is?

A
$$\dfrac{e+1}{\sqrt{e}}$$

B
$$\dfrac{e-1}{\sqrt{e}}$$

C
$$\dfrac{e+1}{2\sqrt{e}}$$

D
$$\dfrac{e-1}{2\sqrt{e}}$$

Solution

given series is $$1+\cfrac{1}{4\times 2!}+\cfrac{1}{16\times 4!}+\cfrac{1}{64\times 6!}+....\infty$$
We know that $${e}^{x}=1+\cfrac{x}{1!}+\cfrac{{x}^{2}}{2!}+\cfrac{{x}^{3}}{3!}+....\infty$$
Let the series be denoted by $$S$$
$$S=1+\cfrac{1}{2!}.{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+\cfrac{1}{4!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 4 }+\cfrac{1}{6!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 6 }+....\infty$$
$$=(1+\cfrac{1}{1!}(\cfrac{1}{2})+\cfrac{1}{2!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+\cfrac{1}{3!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+....\infty)-(\cfrac{1}{1!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+\cfrac{1}{3!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 3 }+....\infty)$$
$$2S=(1+\cfrac{1}{1!}(\cfrac{1}{2})+\cfrac{1}{2!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+.... \infty)+(1-\cfrac{1}{1!}(\cfrac{1}{2})+\cfrac{1}{2!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }-\cfrac{1}{3!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 3 }+....\infty)$$
By $${e}^{x}$$ expansion $${e}^{1/2}=1+\cfrac{1}{1!}(\cfrac{1}{2})+\cfrac{1}{2!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+....\infty)$$
$${e}^{-1/2}=1-\cfrac{1}{1!}(\cfrac{1}{2})+\cfrac{1}{2!}{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }+...\infty$$
$$2S={e}^{1/2}+{e}^{-1/2}$$
$$S=\cfrac { \sqrt { e } +\cfrac { 1 }{ \sqrt { e } } }{ 2 } =\cfrac { e+1 }{ 2\sqrt { e } }
Question 6
1 / -0

The sum of the series $$1+2.2+3.2^{2}+4.2^{3}+5.2^{4}+.+100.2^{99}$$ is ?

A
$$99.2^{100} -1$$

B
$$100.2^{100} +1$$

C
$$99.2^[100}$$

D
$$99.2^{100}+1$$

Solution

$$S=1+2.2+{ 3.2 }^{ 2 }+{ 4.2 }^{ 3 }3+{ 5.2 }^{ 4 }+..+{ 100.2 }^{ 99 }$$

$$2S={ 1 }.{2}^{ 1 }+{ 2.2 }^{ 2 }+{ 3.2 }^{ 3 }+..+{ 99.2 }^{ 99 }+{ 100.2 }^{ 100 }$$

subtracting the both,

$$\left( 1-2 \right) S=1+{ 1.2 }^{ 1 }+{ { 1.2 } }^{ 2 }+{ { 1.2 } }^{ 3 }+..+{ 1.2 }^{ 99 }-{ 100.2 }^{ 100 }$$

$$-S=\left( 1+2+{ 2 }^{ 2 }+{ 2 }^{ 3 }+{ 2 }^{ 4 }+....+{ 2 }^{ 99 } \right) -{ 100.2 }^{ 100 }$$

$$-S=\left( \frac { 1\left( { 2 }^{ 100 }-1 \right) }{ 2-1 } \right) -{ 100.2 }^{ 100 }$$

$$-S={ 2 }^{ 100 }-1-{ 100.2 }^{ 100 }$$\\ $$-S=-1-{ 99.2 }^{ 100 }$$

$$S=1+{ 99.2 }^{ 100 }$$
Question 7
1 / -0

Sum of first n terms of the series $$\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{{15}}{{16}} + ....$$ is equal to

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

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

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

D
none of these

Solution
Question 8
1 / -0

If $${ S }_{ n }=\overset { n }{ \underset { r=1 }{ \Sigma } } { t }_{ r }=\dfrac { 1 }{ 6 } n\left( 2{ n }^{ 2 }+9n+13 \right) $$, then $$\overset { n }{ \underset { r=1 }{ \Sigma } } \sqrt { { t }_{ r } } $$ equals ?

A
$$\dfrac { 1 }{ 2 } n\left( n+1 \right) $$

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

C
$$\dfrac { 1 }{ 2 } n\left( n+3 \right) $$

D
$$\dfrac { 1 }{ 2 } n\left( n+5 \right) $$

Solution
Question 9
1 / -0

The number of zeroes, at the end of $$50!$$, is

A
$$5$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

Solution-
50!$$\rightarrow $$ Number of zeroes of the end
Zeroes can be obtained by multiples of 10
as multiples of 5 and 2.
5 zeroes there are many powers of 2 in 50!
we need to find power of 5 in 50!
Power $$ = 1(5)+1(15)+2(25)+1(35)+1(45)+1(50)$$
$$ = 7 $$
7 zeroes caused by 5, 15, 25, 35, 45, 50
Total zeroes = 5+7 = 12
Question 10
1 / -0

The sum of the series $$1+2(1+1/n)+3(1+1/n)^2+....\infty$$ is given by?

A
$$n^2+1$$

B
$$n(n+1)$$

C
$$n(1+1/n)^2$$

D
$$n^2$$

Solution

$$S = 1+2(1+\cfrac{1}{n}) + 3(1+\cfrac{1}{n})^2+.....\infty$$
$$S(1+\cfrac{1}{n}) = (1+\cfrac{1}{n}) + 2(1+\cfrac{1}{n})^2+.......\infty$$
$$-\cfrac{S}{n} = 1 + (1+\cfrac{1}{n}) + (1+\cfrac{1}{n})^2+......\infty $$
$$-\cfrac{S}{n} = \cfrac{1}{1-(1+\cfrac{1}{n})}$$
$$S = n^2$$