Sequences and Series Test

Result

Sequences and Series Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

If $${ a }_{ n }=\sum _{ r=0 }^{ n }{ \cfrac { 1 }{ { _{ }^{ n }{ C } }_{ r } } } $$, the value of $$\sum _{ r=0 }^{ n }{ \cfrac { n-2r }{ { _{ }^{ n }{ C } }_{ r } } } $$

A
$$\cfrac{n}{2}{a}_{n}$$

B
$$\cfrac{1}{2}{a}_{n}$$

C
$$n{a}_{n}$$

D
$$0$$
Question 2
1 / -0

Observe the pattern carefully
$$11\times11=121$$
$$111\times111=12321$$
$$1111\times1111=\,?$$

A
$$12345321$$

B
$$123421$$

C
$$1234321$$

D
$$12468$$

Solution

$$11\times11=121$$
$$111\times111=12321$$
$$1111\times1111=1234321$$
Question 3
1 / -0

The sum of the series
$$(2)^2+2(4)^{2}+3(6)^{2}+....$$ upto $$10$$ terms is

A
$$12100$$

B
$$11300$$

C
$$11200$$

D
$$12300$$

Solution

$${\left(2\right)}^{2}+2{\left(4\right)}^{2}+3{\left(6\right)}^{2}+…..upto 10 term$$
$${T}_{m}=n{\left(2n\right)}^{2}\Rightarrow 4{n}^{3}$$
$$\sum_{n=0}^{10}{{T}_{n}}=4\sum_{n=0}^{10}{{n}^{3}}=\dfrac{4{\left(n\right)}^{2}{\left(n+1\right)}^{2}}{4}$$
$$n=10\rightarrow =100\times 121=\boxed{12100}$$
Question 4
1 / -0

The sum of the series $$1+\frac { 1.3 }{ 6 } +\frac { 1.3.5 }{ 6.8 } +....\infty$$ is

A
$$1$$

B
$$0$$

C
$$\infty$$

D
$$4$$

Solution

$$ S_n = 1+\dfrac{1.3}{6}+\dfrac{1.3.5}{6.8}+...\infty $$
$$ \Rightarrow 8[\dfrac{1}{2.4}+\dfrac{1.3}{2.4.6}+\dfrac{1.3.5}{2.4.6.8}+]$$
$$ T_{n} = \dfrac{1.3...(2n-1)[(2n+2)-(2n+1)]}{2.4...(2n+2)}$$
$$ T_n = \dfrac{1.3.5...(2n-1)(2n+2)}{2.4.6...(2n+2)}$$
$$ \dfrac{1.3.5...(2n-1)(2n+1)}{2.4...(2n+2)}$$
$$ T_1+T_2 + ... T_n.$$
$$ \dfrac{1.4}{2.4}-\dfrac{1.3}{2.4}+\dfrac{1.3.6}{2.4.6}-\dfrac{1.3.5}{2.4.6}+\dfrac{1.3.5.8}{2.4.6.8}-\dfrac{1.3.5.7}{2.4.6.8}$$
$$ \displaystyle \sum T_n = \dfrac{1.4}{2.4} - \dfrac{1.3.5.7...(2n-1)(2n+1)}{2.4.6.8...(2n+2)} \rightarrow 0 $$
putting $$n\rightarrow \infty $$
$$ \displaystyle \lim_{n\rightarrow \infty } \sum T_n = \dfrac{1}{2}$$
$$ \therefore S_n = 8[\dfrac{1}{2}] = 8\times \dfrac{1}{2} = 4 $$
So, option (d) is correct.
Question 5
1 / -0

If $$A_k=\begin{bmatrix} k & k-1\\ k-1 & k\end{bmatrix}$$ then $$|A_1|+|A_2|+..+|A_{2015}|=?$$

A
$$0$$

B
$$2015$$

C
$$(2015)^2$$

D
$$(2015)^3$$

Solution

$$\begin{array}{l} \left[ { { A_{ 1 } } } \right] =\left| { \left[ { \begin{array} { *{ 20 }{ c } }1 & 0 \\ 0 & 1 \end{array} } \right] } \right| =1 \\ \left[ { { A_{ 2 } } } \right] =\left| { \left[ { \begin{array} { *{ 20 }{ c } }2 & 1 \\ 1 & 2 \end{array} } \right] } \right| =3 \\ \left[ { { A_{ 3 } } } \right] =\left| { \left[ { \begin{array} { *{ 20 }{ c } }3 & 2 \\ 2 & 3 \end{array} } \right] } \right| =5 \\ { A_{ 2015 } }=\left| { \left[ { \begin{array} { *{ 20 }{ c } }{ 2015 } & { 2014 } \\ { 2014 } & { 2015 } \end{array} } \right] } \right| ={ \left( { 2015 } \right) ^{ 2 } }-{ \left( { 2014 } \right) ^{ 2 } }=\left( { 4029 } \right) \\ \therefore \left| { { A_{ 1 } } } \right| \, +\left| { { A_{ 3 } } } \right| +.......\left| { { A_{ 2015 } } } \right| \\ 1+3+5+7....4029 \\ These\, \, series\, \, are\, \, term=1\, common\, difference=2 \\ \Rightarrow 4029\, \, =1+(n-1)\times 2\, \, ,\, \, \frac { { 4028 } }{ 2 } =n-1,n=2015 \\ \Rightarrow Sum=\frac { { 2015 } }{ 2 } \left( { 1+4029 } \right) \\ \Rightarrow \frac { { 2015\times 4030 } }{ 2 } ={ (2015)^{ 2 } } \end{array}$$
Question 6
1 / -0

The value of $$\sum^{10}_{x=1}\sum^{r=x-1}_{r=0}(2^{x}-2^{r})$$ is

A
$$16392$$

B
$$16398$$

C
$$15462$$

D
$$15468$$

Solution

$$\displaystyle\sum_{x=1}^{10}\sum_{r=0}^{x-1}(2^{x}-2^{r})$$
Now, $$\displaystyle\sum_{r=0}^{x-1}(2^{x}-2^{r})=(2^{x}-2^{1})+(2^{x}-2^{2})+......+(2^{x}-2^{x-1})$$
$$=(2^{x}+2^{x}+2^{x}+........+x\ terms)-(2^{0}+2^{1}+2^{2}+......+2^{x-1})$$
$$=x.2^{x}-\left[\dfrac{2^{0}(2^{x}-1)}{2-1}\right]=x.2^{x}-(2^{x}-1)=x.2^{x}-2^{x}+1$$
Now, $$\displaystyle\sum_{x=1}^{10}\sum_{r=0}^{x-1}(2^{c}-2^{r})=\sum_{x=1}^{10}x.2^{x}+\sum_{x=1}^{10}1$$
$$=\displaystyle\sum_{x=1}^{10}x.2^{x}-\left[2^{1}+2^{2}+2^{3}+........+2^{10}\right]+10$$
$$=\displaystyle\sum_{x=1}^{10}x.2^{x}-\left[\dfrac{2(2^{10}-1)}{2-1}\right]+10=\sum_{x=1}^{10}x.2^{x}-2^{11}+12$$
$$=(1.2+2.2^{2}+3.2^{3}+......+10.2^{10})-2^{11}+12$$
$$=18734-2048+12=\boxed{16398}$$ Ans
Question 7
1 / -0

Let $$a=\dfrac{1^{2}}{1}+\dfrac{2^{2}}{3}+\dfrac{3^{2}}{5}+......+\dfrac{(1001)^{2}}{2001}$$, $$b=\dfrac{1^{2}}{3}+\dfrac{2^{2}}{5}+\dfrac{3^{2}}{7}+......+\dfrac{(1001)^{2}}{2003}$$. The closest integer of $$a-b$$ is

A
$$500$$

B
$$501$$

C
$$1000$$

D
$$1001$$

Solution
Question 8
1 / -0

The value of $$1 + \dfrac{x \, \log_e \, 2}{1!} + \dfrac{x^2}{2!} (\log_e 2)^2 + \dfrac{x^3}{3!} (\log_e 2)^3+ ... \infty$$s equal to

A
$$e^2$$

B
$$a^2$$

C
$$2^a$$

D
$$2^x$$

Solution

Series expansion of $$e^{x}$$ is
$$e^{x}=1+\cfrac{x}{1!}+\cfrac{x^{2}}{2!}+\cfrac{x^{3}}{3!}+........\infty $$

So,
$$1+\cfrac{x\log_{e}2}{1!}+\cfrac{\left ( x\log_{e}2 \right )^{2}}{2!}+\cfrac{\left ( x\log_{e}2 \right )^{3}}{3!}+........\infty =e^{x\log_{e}2}$$

$$=e^{\log_{e}2^{x}}$$

$$=\left ( 2^{x} \right )^{\log_{e}e}$$ $$\left ( a^{\log b}=b^{\log a} \right )$$

$$=2^{x}$$
Question 9
1 / -0

If $$z=\dfrac { 1 }{ 3 } +\dfrac { 1.3 }{ 3.6 } +\dfrac { 1.3.5 }{ 3.6.9 } +..........,\infty $$ then

A
$${z}^{2}-2z+2=0$$

B
$${z}^{2}+2z-2=0$$

C
$${z}^{2}-2z-2=0$$

D
$$none\ of\ these$$

Solution
Question 10
1 / -0

Sum of the series
$$(1\times 2015)+(2 \times 2014)+(3\times 2013)......+(2015\times 1)$$ is equal to-

A
$$336\ \times 2015 \times 2016$$

B
$$336\ \times 2015 \times 2017$$

C
$$336\ \times 2016\times 201$$

D
$$None$$

Solution