Sequences and Series Test

Result

Sequences and Series Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

It is known that $$\sum\limits_{r = 1}^\infty {\frac{1}{{{{(2r - 1)}^2}}} = \frac{{{\pi ^2}}}{8},} $$ then $$\sum\limits_{r = 1}^\infty {\frac{1}{{{r^2}}}} $$ is equal to

A
$$\frac{{{\pi ^2}}}{{24}}$$

B
$$\frac{{{\pi ^2}}}{{3}}$$

C
$$\frac{{{\pi ^2}}}{{6}}$$

D
none of these

Solution

Given : $$\displaystyle \sum_{r = 1}^{\infty } \frac{1}{(2r-1)^{2}} = \frac{\pi ^{2}}{8}, \sum_{r = 1}^{\infty } \frac{1}{r^{2}} = ?? $$
start putting 'r' value
$$\displaystyle \frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}------ = \frac{\pi ^{2}}{8} $$
let say
$$\displaystyle \sum_{r = 1}^{\infty } \frac{1}{r^{2}} = k $$
$$\displaystyle \frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+-----\frac{1}{r^{2}} = k $$
$$\displaystyle [\frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+-----\frac{1}{(2r-1)^{2}}]+\frac{1}{2^{2}}+\frac{1}{4^{2}}+---- = k $$
$$\displaystyle \frac{\pi ^{2}}{8}+\frac{1}{2^{2}}[\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}----] = k $$
$$\displaystyle \frac{\pi ^{2}}{8}+\frac{1}{4}(k) = k $$
$$\displaystyle \frac{\pi ^{2}}{8} = k-\frac{k}{4} = \frac{3k}{4} $$
so, $$\displaystyle k = \frac{\pi ^{2}}{6} $$ Answer option- C
Question 2
1 / -0

The sum $$\frac{3}{{1.2}},\frac{3}{{1.2}},\frac{1}{2},\frac{4}{{2.3}}{\left( {\frac{1}{2}} \right)^2} + \frac{5}{{3.4}}{\left( {\frac{1}{2}} \right)^2}$$

A
$$1 - \frac{1}{{\left( {n + 1} \right){2^n}}}$$

B
$$1 - \frac{1}{{n{{.2}^{n - 1}}}}$$

C
$$1 = \frac{1}{{\left( {n + 1} \right){2^n}}}$$

D
$$\frac{1}{{\left( {n - 1} \right){2^{n - 1}}}}$$

Solution

$$t_n=\dfrac{n+2}{n(n+1)}\cdot \left(\dfrac{1}{2}\right)^n$$

$$=\dfrac{2(n+1)-n}{n(n+1)} \cdot \left(\dfrac{1}{2}\right)^n$$

$$t_n=\dfrac{1}{n}\left(\dfrac{1}{2}\right)^{n-1}-\dfrac{1}{n+1} \cdot \left(\dfrac{1}{2}\right)^n$$

$$S_n=\sum_{n=1}^nt_n$$

$$=\left\{\dfrac{1}{1}\left(\dfrac{1}{2}\right)^0-\dfrac{1}{2}\left(\dfrac{1}{2}\right)^1\right\}$$ $$+\left\{\dfrac{1}{2}\left(\dfrac{1}{2}\right)^1-\dfrac{1}{3}\left(\dfrac{1}{2}\right)^2\right\}+...+$$ $$=\left\{\dfrac{1}{n}\left(\dfrac{1}{2}\right)^{n-1}-\dfrac{1}{n+1}\left(\dfrac{1}{2}\right)^n\right\}$$

$$S_n=1-\dfrac{1}{(n+1)2^n}$$
Question 3
1 / -0

If f(x)= $$x+\frac{1}{2x+\frac{1}{\frac{1}{2x+...\infty }}}$$ ;
then the value of f (2011). f'(2011) is :

A
0

B
1

C
2011

D
2010

Solution
Question 4
1 / -0

Equation $${x^n} - 1 = 0,n \in N,$$ has roots $$1,{a_2},.....{a_n}$$.
The value of $$\sum\limits_{r = 2}^n {\frac{1}{{1 - {a_r}}},} $$ is

A
$$\frac{n}{4}$$

B
$$\frac{{n\left( {n - 1} \right)}}{2}$$

C
$$\frac{{\left( {n - 1} \right)}}{2}$$

D
none of these

Solution
Question 5
1 / -0

Fill in the missing values.
J V R B
Z 18 24 22 14
L 11 17 15 7
? 9 15 13 5
T 15 21 ? 11

A
X and $$29$$

B
P and $$15$$

C
H and $$19$$

D
F and $$12$$

Solution
Question 6
1 / -0

The number of the three digits numbers having only two consecutive digits identical is:

A
153

B
160

C
180

D
161

Solution
Question 7
1 / -0

$$1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + \frac{1}{{1 + 2 + 3 + 4}} + ...... + \frac{1}{{1 + 2 + ... + 2015}}$$

A
$$\frac{{4030}}{{2016}}$$

B
$$\frac{{2030}}{{2016}}$$

C
$$\frac{{3030}}{{2016}}$$

D
$$\frac{{3015}}{{2016}}$$

Solution
Question 8
1 / -0

Find the missing number in the given figure?

A
$$25$$

B
$$361$$

C
$$196$$

D
$$144$$

Solution
Question 9
1 / -0

If $$24 + 37 = 7,12 + 18 = 3$$, then 54+21=?

A
11

B
9

C
12

D
3

Solution
Question 10
1 / -0

In the following number series,one number is wrong.find out the ?

1,2,8,33,149,765,4626

A
$$33$$

B
$$8$$

C
$$149$$

D
$$0$$

Solution

1,2,8,33,149,765,4626
Carefully analyzing the series, we can see the following pattern
$$1\times 1+(1)^{2}=2$$
$$2\times 2+(2)^{2}=8$$
$$8\times 3+(3)^{2}=33$$
$$33\times 4+(4)^{2}=148$$
$$148\times 5+(5)^{2}=765$$
$$765\times 6+(6)^{2}=4626$$
So, we can clearly see that 149 is the wrong entry in the series.