Sequences and Series Test

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Sequences and Series Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle \sum_{s=1}^{n}\, \left \{ \displaystyle \sum_{r=1}^{s}r \right \}\, =\, an^3\, +\, bn^2\, +\, cn$$, then find the value of a + b + c.

A
1

B
0

C
2

D
3

Solution

$$\displaystyle \sum_{s=1}^{n}\, \left \{ \displaystyle \sum_{r=1}^{s}r \right \}\, =\, \displaystyle \sum_{s=1}^{n}\, \displaystyle \frac {s(s+1)}{2}$$

$$=\, \displaystyle \frac {1}{2}\, \left [ \displaystyle \frac {n(n+1)(2n+1)}{6}\, +\, \displaystyle \frac {n(n+1)}{2} \right ]$$

$$=\, \displaystyle \frac {n(n+1)}{4}\, \left [ \displaystyle \frac {2n+1+3}{3} \right ]\, =\, \displaystyle \frac {1}{12}\, n(n+1)(2n+4)$$

$$\Rightarrow\, a\, =\, \displaystyle \frac {1}{6},\, b\, =\, \displaystyle \frac {1}{2},\, c\, =\, \displaystyle \frac {1}{3}$$

$$\Rightarrow a+b+c=1$$
Question 2
1 / -0

The sum of n terms of the series
$$1 + (1 + a) + (1 + a + a^{2}) + (1 + a + a^{2} + a^{3}) + .....$$, is

A
$$\displaystyle \frac{n}{1 - a} - \frac{a(1 - a^{n})}{(1 - a)^{2}}$$

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

C
$$\displaystyle \frac{n}{1 - a} - \frac{a(1 + a^{n})}{(1 - a)^{2}}$$

D
$$\displaystyle - \frac {n}{1 - a} + \frac{a(1 - a^{n})}{(1 - a)^{2}}$$

Solution

$$T_n = 1 + a + a^{2} + ..... + a^{n - 1} = \displaystyle \frac{1 - a^{n}}{1 - a}$$

$$S_n = \Sigma T_n =\displaystyle \frac{1}{(1 -a)} [\Sigma 1 - \Sigma a^{n}]$$

$$= \displaystyle \frac{1}{(1 - a)} [n - (a + a^{2} + a^{3} + ..... a^{n})]$$

$$= \displaystyle \frac{1}{(1 - a)} \left[n - \frac{a(1 - a^{n})}{(1 - a)} \right] = \frac{n}{1 - a} - \frac{a(1 - a^{n})}{(1 - a)^{2}}$$
Question 3
1 / -0

The sum of series $$\displaystyle \frac {1^2}{1}+\frac {1^2+2^2}{1+2}+\frac {1^2+2^2+3^2}{1+2+3}+....$$ upto n terms is

A
$$\displaystyle\frac {1}{3}(2n+1)$$

B
$$\displaystyle\frac {1}{3}n^2$$

C
$$\displaystyle\frac {1}{3}(n+2)$$

D
$$\displaystyle\frac {1}{3}n(n+2)$$

Solution

$$\displaystyle S=\frac { { 1 }^{ 2 } }{ 1 } +\frac { { 1 }^{ 2 }+{ 2 }^{ 2 } }{ 1+2 } +\frac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 1+2+3 } +...$$

$$\Rightarrow\displaystyle S=\sum _{ r=1 }^{ n }{ \frac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ r }^{ 2 } }{ 1+2+3+...+r } } $$

$$\displaystyle=\sum _{ r=1 }^{ n }{ \frac { \displaystyle\frac { r\left( r+1 \right) \left( 2r+1 \right) }{ 6 } }{ \displaystyle\frac { r\left( r+1 \right) }{ 2 } } } =\sum _{ r=1 }^{ n }{ \left( \displaystyle\frac { 2r+1 }{ 3 } \right) } $$

$$\displaystyle=\frac { 2 }{ 3 } \sum _{ r=1 }^{ n }{ r } +\frac { 1 }{ 3 } \sum _{ r=1 }^{ n }{ 1 } =\frac { 2 }{ 3 } \frac { n(n+1) }{ 2 } +\frac { 1 }{ 3 } n$$

$$\displaystyle =\frac { n }{ 3 } \left( n+1+1 \right) =\frac { n }{ 3 } \left( n+2 \right) $$
Question 4
1 / -0

$$\dfrac {1^3+2^3+3^3+4^3+........12^3}{1^2+2^2+3^2+4^2+......12^2}=$$

A
$$\dfrac {234}{25}$$

B
$$\dfrac {243}{35}$$

C
$$\dfrac {263}{27}$$

D
None of these

Solution

Clearly required ratio is $$=\cfrac{\sum n^3}{\sum n^2}=\cfrac{\cfrac{n^2(n+1)^2}{4}}{\cfrac{ n(n+1) (2n+1)}{6}}=\cfrac{3n(n+1)}{2(2n+1)}$$
Substituting $$n =12$$ we get required value $$=\cfrac{234}{25}$$
Hence, option 'A' is correct.
Question 5
1 / -0

1 + 3 + 6 + 10 + .... upto n terms is equal to

A
$$\displaystyle \frac{1}{3} n(n + 1)(n + 2)$$

B
$$\displaystyle \frac{1}{6} n(n + 1)(n + 2)$$

C
$$\displaystyle \frac{1}{12} n(n + 2)(n + 3)$$

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

Solution

Let $$S =1+3+6+10+..............+T_n$$ ........(1)
Also $$S = 1+3 + 6+ 10+..........+T_{n-1} +T_n$$ ........(2)
Now by $$(1)-(2)$$
$$0=1+2+3+.........+n-T_n$$
$$\Rightarrow T_n = 1+2+3......+n=\cfrac{1}{2}n(n+1)$$
Hence required summation is,
$$\displaystyle \sum_{k=1}^nT_k=\frac{1}{2}\left(\sum_{k=1}^nk^2+\sum_{k=1}^n k \right)$$
$$\displaystyle =\frac{1}{2}\left(\frac{1}{6}n(n+1)(2n+1)+\frac{1}{2}n(n+1) \right)$$
$$\quad =\displaystyle \frac{1}{6}n(n+1)(n+2)$$
Question 6
1 / -0

The value of $$\displaystyle \sum_{n=1}^{10}\left (\sin \dfrac {2n\pi}{11}-\cos\dfrac {2n\pi}{11}\right )$$ is equal to

A
$$2$$

B
$$1$$

C
$$0$$

D
$$-1$$

Solution

Given series
$$=\left (\sin \dfrac {2\pi}{11}-\cos\dfrac {2\pi}{11}\right )+\left (\sin \dfrac {4\pi}{11}-\cos\dfrac {4\pi}{11}\right )+\left (\sin \dfrac {6\pi}{11}-\cos\dfrac {6\pi}{11}\right )+....+\left (\sin \dfrac {20\pi}{11}-\cos\dfrac {20\pi}{11}\right )$$
$$=\left (\sin \dfrac {2\pi}{11}+\sin\dfrac {4\pi}{11}+....+\sin\dfrac {20\pi}{11}\right )$$$$-\left (\cos \dfrac {2\pi}{11}+\cos\dfrac {4\pi}{11}+....+\cos\dfrac {20\pi}{11}\right )$$
$$=\dfrac {\sin \pi.\sin \dfrac {10.\pi}{11}}{\sin\dfrac {\pi}{11}}-\dfrac {\cos \pi.\sin \dfrac {10\pi}{11}}{\sin\dfrac {\pi}{11}}$$
$$=0+\dfrac {\sin\left (\pi-\dfrac {\pi}{11}\right )}{\sin \frac {\pi}{11}}=1$$
Question 7
1 / -0

The sum of the first n terms of the series $$\displaystyle 1^{2}+2.2^{2}+3^{3}+2.4^{2}+5^{2}+2.6^{2}+....$$ is $$\displaystyle \frac{n\left ( n+1 \right )^{2}}{2}$$ when n is even. When n is odd the sum is

A
$$\displaystyle \frac{n\left ( n+1 \right )^{2}}{4}$$

B
$$\displaystyle \frac{n^{2}\left ( n+2 \right )}{4}$$

C
$$\displaystyle \frac{n^{2}\left ( n+1 \right )}{4}$$

D
$$\displaystyle \frac{n\left ( n+2 \right )^{2}}{4}$$

Solution

if $$n$$ is even then sum $$= \dfrac{n(n+1)^2}{2}$$
if $$n$$ is odd, last term $$= n^2$$
required sum

$$= [1^2+ 2.2^2 + 3^2+ 2.4^2 +….. + 2(n-1)^2] + n^2$$

$$= \dfrac{(n-1)n^2}{2} +n^2$$ [ Using sum for $$( n – 1 )$$ terms to be even $$=\dfrac{(n-1)n^2}{2}$$ ]
required sum $$=\dfrac {n^2(n+1)}{2}$$
Question 8
1 / -0

The infinite sum $$1\, +\, \displaystyle {\frac{4}{7}\, +\, \frac{9}{7^2}\, +\, \frac{16}{7^3}\, +\, \frac{25}{7^4}\, +\, ......}$$ equals

A
$$\displaystyle \frac{27}{14}$$

B
$$\displaystyle \frac{21}{13}$$

C
$$\displaystyle \frac{49}{27}$$

D
$$\displaystyle \frac{256}{147}$$

Solution

Let $$x = 1+\displaystyle \frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+\frac{25}{7^4}+..$$
Dividing both the sides by 7
$$\displaystyle \frac{x}{7} = \frac{1}{7}+\frac{4}{7^2}+\frac{9}{7^3}+\frac{16}
{7^4}+...$$
Subtracting iind from the 1st
$$\displaystyle \frac{6x}{7} = 1+\frac{3}{7}+\frac{5}{7^2}+\frac{7}{7^3}+\frac{9}{7^4}...$$
The RHS here is an arithmetic-Geometric series , Now divide the equation by 7 again
$$\displaystyle \frac{6x}{49} = \frac{1}{7}+\frac{3}{7^2}+\frac{5}{7^3}+\frac{7}{7^4}$$
Again subtracting if from the second one
$$\Rightarrow \displaystyle \frac{36x}{49}-1 = \frac{\frac{2}{7}}{1-\frac{1}{7}}$$
$$\Rightarrow \displaystyle \frac{36x}{49}-1 = \frac{\frac{2}{7}}{\frac{6}{7}}$$
$$\Rightarrow \displaystyle \frac{36x}{49} = 1+\frac{1}{3} = \frac{4}{3}$$

$$\therefore x= \displaystyle \frac{49}{27}$$
Question 9
1 / -0

If $$\displaystyle \sum_{r=1}^{n}t_{r}=\frac{1}{12}n\left ( n+1 \right )\left ( n+2 \right )$$, then the value of $$\displaystyle \sum_{r=1}^{n}\frac{1}{t_{r}}$$ is

A
$$\displaystyle \frac{2n}{n+1}$$

B
$$\displaystyle \frac{n}{\left ( n+1 \right )}$$

C
$$\displaystyle \frac{4n}{ n+1 }$$

D
$$\displaystyle \frac{3n}{ n+1 }$$

Solution

If we take $${ t }_{ r }=\cfrac { 1 }{ 4 } n(n+1)$$
$${ t }_{ r }=\cfrac { 1 }{ 4\times 3 } n(n+1)[(n+2)-(n-1)]=\cfrac { 1 }{ 12 } [n(n+1)(n+2)-n(n+1)(n-1)]\\ { t }_{ 1 }=\cfrac { 1 }{ 12 } [1.2.3-0]\\ { t }_{ 2 }=\cfrac { 1 }{ 12 } [2.3.4-1.2.3]\\ { t }_{ 3 }=\cfrac { 1 }{ 12 } [3.4.5-2.3.4]\\ \underline { { t }_{ n }=\cfrac { 1 }{ 12 } [n(n+1)(n+2)-n(n+1)(n-1)] } \\ { \sum }_{ r=1 }^{ n }{ t }_{ r }=\cfrac { 1 }{ 12 } [n(n+1)(n+2)]\quad \quad Hence,\quad { t }_{ r }=\cfrac { n(n+1) }{ 4 } \\ \cfrac { 1 }{ { t }_{ r } } =\cfrac { 4 }{ n(n+1) } =4\left[ \cfrac { (n+1)-n }{ n(n+1) } \right] =4\left[ \cfrac { 1 }{ n } -\cfrac { 1 }{ (n+1) } \right] \\ \cfrac { 1 }{ { t }_{ 1 } } =4\left[ 1-\cfrac { 1 }{ 2 } \right] \\ \cfrac { 1 }{ { t }_{ 2 } } =4\left[ \cfrac { 1 }{ 2 } -\cfrac { 1 }{ 3 } \right] \\ \underline { \cfrac { 1 }{ { t }_{ n } } =4\left[ \cfrac { 1 }{ n } -\cfrac { 1 }{ (n+1) } \right] \quad \quad \quad } \\ { \sum }_{ r=1 }^{ n }\cfrac { 1 }{ { t }_{ r } } =4\left[ 1-\cfrac { 1 }{ (n+1) } \right] =\cfrac { 4n }{ (n+1) } $$
Question 10
1 / -0

If $$\displaystyle x_{i}> 0,i=1,2,....,50\: \: and\: \: x_{1}+x_{2}+...+x_{50}=50 $$ then the minimum value of $$\displaystyle \frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{50}}$$ equals to

A
$$50$$

B
$$\displaystyle \left ( 50 \right )^{2}$$

C
$$\displaystyle \left ( 50 \right )^{3}$$

D
$$\displaystyle \left ( 50 \right )^{4}$$

Solution

Using $$AM\ge HM$$, we get
$$\dfrac{x_1+x_2+x_3+.......+x_{50}}{50}\ge \dfrac{50}{\dfrac{1}{x_1}+\dfrac{1}{x_2}+............+\dfrac{1}{x_{50}}}$$
$$\Rightarrow \dfrac{50}{50}\ge \dfrac{50}{\dfrac{1}{x_1}+\dfrac{1}{x_2}+............+\dfrac{1}{x_{50}}}$$, use the given values

$$\therefore \dfrac{1}{x_1}+\dfrac{1}{x_2}+........+\dfrac{1}{x_{50}}\ge 50$$

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Re-Attempt Weekly Quiz Competition

Sequences and Series Test - 22

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Sequences and Series Test - 22

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SHARING IS CARING

Question 1

1

0

2

3

Solution

Question 2

$$\displaystyle \frac{n}{1 - a} - \frac{a(1 - a^{n})}{(1 - a)^{2}}$$

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

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

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

Solution

Question 3

$$\displaystyle\frac {1}{3}(2n+1)$$

$$\displaystyle\frac {1}{3}n^2$$

$$\displaystyle\frac {1}{3}(n+2)$$

$$\displaystyle\frac {1}{3}n(n+2)$$

Solution

Question 4

$$\dfrac {234}{25}$$

$$\dfrac {243}{35}$$

$$\dfrac {263}{27}$$

None of these

Solution

Question 5

$$\displaystyle \frac{1}{3} n(n + 1)(n + 2)$$

$$\displaystyle \frac{1}{6} n(n + 1)(n + 2)$$

$$\displaystyle \frac{1}{12} n(n + 2)(n + 3)$$

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

Solution

Question 6

$$2$$

$$1$$

$$0$$

$$-1$$

Solution

Question 7

$$\displaystyle \frac{n\left ( n+1 \right )^{2}}{4}$$

$$\displaystyle \frac{n^{2}\left ( n+2 \right )}{4}$$

$$\displaystyle \frac{n^{2}\left ( n+1 \right )}{4}$$

$$\displaystyle \frac{n\left ( n+2 \right )^{2}}{4}$$

Solution

Question 8

$$\displaystyle \frac{27}{14}$$

$$\displaystyle \frac{21}{13}$$

$$\displaystyle \frac{49}{27}$$

$$\displaystyle \frac{256}{147}$$

Solution

Question 9

$$\displaystyle \frac{2n}{n+1}$$

$$\displaystyle \frac{n}{\left ( n+1 \right )}$$

$$\displaystyle \frac{4n}{ n+1 }$$

$$\displaystyle \frac{3n}{ n+1 }$$

Solution

Question 10

$$50$$

$$\displaystyle \left ( 50 \right )^{2}$$

$$\displaystyle \left ( 50 \right )^{3}$$

$$\displaystyle \left ( 50 \right )^{4}$$

Solution

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