Sequences and Series Test

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Sequences and Series Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

If $$(1+x+x^{2})^{n}=a_{0}+a_{1}+a_{2}x^{2n}$$, then $$a_{0}+a_{2}+a_{4}+........+a_{2n}x^{2n},$$ is equal to :

A
$$\dfrac{3^{n}+1}{2}$$

B
$$\dfrac{3^{n}-1}{2}$$

C
$$\dfrac{1-3^{n}}{2}$$

D
$$3^{n}+\dfrac{1}{2}$$

Solution

$$ { \left( { 1+x+{ x^{ 2 } } } \right) ^{ 4 } }={ a_{ 0 } }+{ a_{ 1 } }+{ a_{ 2 } }{ x^{ 2n } }+.........{ x^{ 4 } }+........{ a_{ 2x } }{ x^{ 2n } }\, \, \, \to \left( i \right) $$

$$\\ Put\, \, x=1\, \, in\, \, equation\, \left( i \right) \\ { 3^{ x } }={ a_{ 0 } }+{ a_{ 1 } }+{ a_{ 2 } }+....................\to \left( { ii } \right) $$

$$\\ Put\, \, x=-1\, \, \, in\, \, equation\, \, \left( i \right) \\ 1={ a_{ 0 } }-{ a_{ 1 } }+{ a_{ 2 } }-{ a_{ 3 } }+..............\to \left( { iii } \right) \\ $$

$$ On\, adding\, \, equation\, \, \left( { ii } \right) \, \, and\, \, \, \left( { iii } \right) \\ { 3^{ n } }\, +1=2\left( { { a_{ 0 } }+{ a_{ 2 } }+{ a_{ 4 } }+......... } \right) \\ \dfrac{ { 3^{ n } }+1}{2}={ a_{ 0 } }+{ a_{ 2 } }+{ a_{ 4 } }+............$$

Hence, this is the answer.
Question 2
1 / -0

In the sum $$3+33+333+3333+.........2015$$ terms the number formed by taking the last four digits in that order is

A
$$6365$$

B
$$6255$$

C
$$6465$$

D
$$6565$$

Solution
Question 3
1 / -0

$$^{(2n+1)}C_{0}+^{(2n+1)}C_{1}+^{(2n+1)}C_{2}+...+^{(2n+1)}C_{n}=$$

A
$$2^{n}$$

B
$$2^{-n}$$

C
$$2^{2n}$$

D
$$3^{2n}$$

Solution
Question 4
1 / -0

Sum of the series $$\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\dfrac{3^2}{3!}+\dfrac{4^2}{4!}+.....$$(infinite terms) is?

A
e

B
$$2e$$

C
$$e^2$$

D
$$\infty$$

Solution

$$ \begin{aligned} &S=\frac{1^{2}}{1 !}+\frac{2^{2}}{2 !}+\frac{3^{2}}{3 !}+\frac{4^{2}}{4 b}+\\ &\text {(infinite terems)} is \\ &S=\sum_{n=1}^{\infty} n^{2}\\ &n^{2}=a_{0}+a_{1} n+a_{2} n(n-1)\\ &n=0 \quad 0=a_{0}\\ &n=1 \quad 1=0+a, \Rightarrow a_{1}=1\\ &n=2 \quad 4=0+2+2 a_{2} \Rightarrow a_{2}=1\\ &S=\sum_{n=1}^{\infty}\left(a_{0}+a_{1} n+a_{2} n(n-1)\right) \end{aligned} $$
$$S=\sum_{n=1}^{\infty}\left(\frac{0+n+n(n-1)}{n_{0}}\right)$$ $$S=\sum_{n=1}^{\infty} \frac{n}{n_{b}}+\sum_{n=1}^{\infty} \frac{n(n-1)}{n_{0}^{1}}$$
$$S=e+e=2 e$$
$$e=1+\frac{1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots} \Rightarrow \sum_{n=1}^{\infty} \frac{1}{(n-1) !}$$ $$e=\sum_{n=2(n-2)\}}=1+1+\frac{1}{2 !}+\frac{1}{3 !} \cdots \cdot \cdots$$
Option $$B$$ is correct
Question 5
1 / -0

I for $$n\in I, n> I0;1+(1+x)+(1+x)^{2}+....+(1+)^{n}=\displaystyle \sum_{k=0}^{n}a_{k}.x^{k},x\neq 0$$ then

A
$$\displaystyle \sum_{k=0}^{n}a_{k}=2^{n+1}$$

B
$$a_{n-2}=\dfrac {n(n+1)}{2}$$

C
$$a_{p}> a_{p-1}$$ for $$p< \dfrac {n}{2}, p\in N$$

D
$$(a_{9})^{2}-(a_{8})^{2}=^{n+2}C_{10}(^{n+1}C_{1}-^{n+1}C_{9})$$

Solution
Question 6
1 / -0

If $$(3x-1)^{}=a_{7}x^{7}+a_{6}x^{6}+....+a_{0}$$, then $$a_{7}+a_{6}+...+a_{0}$$ is

A
$$128$$

B
$$1$$

C
$$64$$

D
$$None\ of\ these$$

Solution

$$\begin{array}{l} { \left( { 3x-1 } \right) ^{ 7 } }=\left[ { ^{ 7 }{ C_{ 0 } }.{ { \left( { 3x } \right) }^{ 7 } }{ -^{ 7 } }{ C_{ 1 } }{ { \left( { 3x } \right) }^{ 6 } }{ +^{ 7 } }{ C_{ 2 } }{ { \left( { 3x } \right) }^{ 5 } }{ +^{ 7 } }{ C_{ 3 } }{ { \left( { 3x } \right) }^{ 5 } }{ +^{ 7 } }{ C_{ 4 } }{ { \left( { 3x } \right) }^{ 4 } }{ -^{ 7 } }{ C_{ 5 } }{ { \left( { 3x } \right) }^{ 5 } }{ +^{ 7 } }{ C_{ 6 } }{ { \left( { 3x } \right) }^{ 6 } }{ -^{ 7 } }{ C_{ 7 } }{ { \left( { 3x } \right) }^{ 7 } } } \right] \\ \Rightarrow 2187{ x^{ 7 } }-5103{ x^{ 6 } }+5103{ x^{ 5 } }-2035{ x^{ 4 } }+945{ x^{ 3 } }-189{ x^{ 2 } }+21x-1\, \, \, \, \, \, \left[ \begin{array}{l} { ^{ n } }{ C_{ 1 } }=n \\ ^{ n }{ C_{ 0 } }{ =^{ n } }{ C_{ n } }=1 \end{array} \right] \\ \Rightarrow \therefore { a_{ 7 } }+{ a_{ 6 } }+{ a_{ 5 } }+{ a_{ 4 } }+............{ a_{ 0 } }=2187-5103+5103-2835+945-189+21-1=128\, \, \, \end{array}$$
Question 7
1 / -0

$$\dfrac { 7 } { 11 } : \dfrac { 336 } { 110 } : ? \quad : \quad \dfrac { 720 } { 272 }$$

A
$$\dfrac { 9 } { 17 }$$

B
$$\dfrac { 9 } { 13 }$$

C
$$\dfrac { 11 } { 13 }$$

D
$$\dfrac { 11 } { 17 }$$

Solution
Question 8
1 / -0

Find the number .

A
$$473$$

B
$$623$$

C
$$389$$

D
$$584$$

Solution

$$ \begin{array}{l} 4^{3}+7^{3}+3^{3}=434 \\ \text { similarly, } \\ 6^{3}+5^{3}+2^{3}=349 \\ \Rightarrow \text { ? }=8^{3}+4^{3}+2^{3} \\ \Rightarrow \text { ? }=584 \\ \text { option } D \text { is correct. } \end{array} $$
Question 9
1 / -0

$$ABCDEFGHIJKLMZYXWVUTSRQPON$$

From the above letter series find the letter which is at the $$6th$$ position to the right side of the letter which is at the centre position of the letters which are at the $$11th$$ place form the left and $$14th$$ place from the right.

A
$$V$$

B
$$F$$

C
$$Y$$

D
$$L$$

Solution
Question 10
1 / -0

There is a specific relationship between the numbers that are given in the following figures. On the basis of the relationship choose the correct alternative to replace the question mark.

A
$$210$$

B
$$266$$

C
$$288$$

D
$$318$$