Sequences and Series Test

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Sequences and Series Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

Let x, y, z be three positive prime numbers. The progression in which $$\sqrt{x}$$, $$\sqrt{y}$$, $$\sqrt{z}$$ can be three terms (not necessarily consecutive) is

A
AP

B
GP

C
HP

D
none of these

Solution

$$ x,y,z $$ are primes $$\Rightarrow \sqrt{x},\sqrt{y},\sqrt{z} $$ are irrationals

Let $$\sqrt{x}=a+pd; \sqrt{y}=a+qd; \sqrt{z}=a+rd $$

Now, $$(\sqrt{y}-\sqrt{x})^{2} = ((q-p)d)^{2} $$

$$ \Rightarrow x+y-2\sqrt{xy}=((q-p)d)^{2} $$

LHS is irrational whereas RHS is rational

Not Possible

Hence, Not in AP

Let $$\sqrt{x}=ar^{p}; \sqrt{y}=ar^{q}; \sqrt{z}=ar^{s} $$

$$\therefore \dfrac{\sqrt{y}}{\sqrt{x}} = r^{q-p} $$

$$\Rightarrow y= x(r^{2(q-p)})$$

$$\Rightarrow y$$ has two factors $$ x $$ and $$r^{2(q-p)}$$

Not possible as $$y$$ is prime

Hence, not in GP

Let $$\dfrac{1}{\sqrt{x}}=a+pd; \dfrac{1}{\sqrt{y}}=a+qd; \dfrac{1}{\sqrt{z}}=a+rd $$

$$\dfrac{1}{\sqrt{y}}-\dfrac{1}{\sqrt{x}} = (q-p)d$$

$$ \Rightarrow \dfrac{(\sqrt{y}-\sqrt{x})^{2}}{yx} = ((p-q)d)^{2} $$

$$ x+y-2\sqrt{yx} = xy((p-q)d)^{2} $$

LHS = irrational and RHS = rational

Not possible. Hence, Not in HP
Question 2
1 / -0

Let $$\displaystyle a_{1},a_{2},a_{3},....,a_{11}$$ be real numbers satisfying $$\displaystyle a_{1}=15,27-2a_{2}> 0\:$$ and $$\: \: a_{k}=2a_{k-1}-a_{k-2}$$ for $$k = 3, 4, ......., 11$$.
If $$\displaystyle \frac{a_{1}^{2}+a_{2}^{2}+...+a_{11}^{2}}{11}=90 $$, then the value of $$\displaystyle \frac{a_{1}+a_{2}+...+a_{11}}{11} $$ is equal to

A
1

B
5

C
9

D
0

Solution

$${ a }_{ 1 }=15\\ { a }_{ 3 }=2{ a }_{ 2 }-{ a }_{ 1 }\\ { a }_{ 4 }=2{ a }_{ 3 }-{ a }_{ 2 }=4{ a }_{ 2 }-2{ a }_{ 1 }-{ a }_{ 2 }=3{ a }_{ 2 }-2{ a }_{ 1 }\\ { a }_{ 5 }=2{ a }_{ 4 }-{ a }_{ 3 }=6{ a }_{ 2 }-4{ a }_{ 1 }-2{ a }_{ 2 }+{ a }_{ 1 }=4{ a }_{ 2 }-3{ a }_{ 1 }\\ { a }_{ 6 }=5{ a }_{ 2 }-6{ a }_{ 1 }-3{ a }_{ 1 }+2{ a }_{ 1 }=5{ a }_{ 2 }-4{ a }_{ 1 }\\ { a }_{ 7 }=2{ a }_{ 6 }-{ a }_{ 5 }=10{ a }_{ 2 }-8{ a }_{ 1 }-4{ a }_{ 2 }+3{ a }_{ 1 }=6{ a }_{ 2 }-5{ a }_{ 1 }\\ { a }_{ k }=(k-1){ a }_{ 2 }-(k-2){ a }_{ 1 }\quad \quad k\ge 3\\ \cfrac { { { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+{ { a }_{ 3 } }^{ 2 }+....+{ { a }_{ 11 } }^{ 2 } }{ 11 } =90$$
$$={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+\overset { 11 }{ \underset { k=3 }{ \sum   } } \left[ { (k-1) }^{ 2 }{ { a }_{ 2 } }^{ 2 }+{ (k-2) }^{ 2 }{ { a }_{ 1 } }^{ 2 }-2(k-1)(k-2){ a }_{ 1 }{ a }_{ 2 } \right] =990\\ ={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }\overset { 11 }{ \underset { k=3 }{ \sum   } } \left( { k }^{ 2 }+1-2k \right) +\overset { 11 }{ \underset { k=3 }{ \sum   } } \left( { k }^{ 2 }+4-4k \right) { { a }_{ 1 } }^{ 2 } \\ -2{ a }_{ 1 }{ a }_{ 2 }\overset { 11 }{ \underset { k=3 }{ \sum   } } \left( { k }^{ 2 }-3k+2 \right) =990$$

$$={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+\overset { 11 }{ \underset { k=3 }{ \sum   } } \left( { { a }_{ 2 } }^{ 2 }+{ { a }_{ 1 } }^{ 2 }-2{ a }_{ 1 }{ a }_{ 2 } \right) { k }^{ 2 }+\overset { 11 }{ \underset { k=3 }{ \sum   } } k\left( -2{ { a }_{ 2 } }^{ 2 }-4{ { a }_{ 1 } }^{ 2 }+6{ a }_{ 1 }{ a }_{ 2 } \right) \\ +\left( { { a }_{ 2 } }^{ 2 }+4{ { a }_{ 1 } }^{ 2 }-4{ a }_{ 1 }{ a }_{ 2 } \right) \overset { 11 }{ \underset { k=3 }{ \sum   } } (1)=990\\ ={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+\left( { { a }_{ 2 } }^{ 2 }+{ { a }_{ 1 } }^{ 2 }-2{ a }_{ 1 }{ a }_{ 2 } \right) { \left[ \cfrac { k(k+1)(2k+1) }{ 6 } \right] }_{ 2 }^{ 11 }+\left( -2{ { a }_{ 2 } }^{ 2 }-4{ { a }_{ 1 } }^{ 2 }+6{ a }_{ 1 }{ a }_{ 2 } \right) { \left[ \cfrac { k(k+1) }{ 2 } \right] }_{ 2 }^{ 11 } \\ +\left( { { a }_{ 2 } }^{ 2 }+4{ { a }_{ 1 } }^{ 2 }-4{ a }_{ 1 }{ a }_{ 2 } \right) { \left[ k \right] }_{ 2 }^{ 11 }=990$$

$$={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+\left( { { a }_{ 2 } }^{ 2 }+{ { a }_{ 1 } }^{ 2 }-2{ a }_{ 1 }{ a }_{ 2 } \right) \left[ \cfrac { 11(12)(23)-2(3)(5) }{ 6 } \right] \\ +\left( -2{ { a }_{ 2 } }^{ 2 }-4{ { a }_{ 1 } }^{ 2 }+6{ a }_{ 1 }{ a }_{ 2 } \right) \left[ \cfrac { 11(12)-2(3) }{ 2 } \right] +\left( { { a }_{ 2 } }^{ 2 }+4{ { a }_{ 1 } }^{ 2 }-4{ a }_{ 1 }{ a }_{ 2 } \right) \left[ 11-2 \right] =990$$

$$={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+\left( { { a }_{ 2 } }^{ 2 }+{ { a }_{ 1 } }^{ 2 }-2{ a }_{ 1 }{ a }_{ 2 } \right) (501)+\left( -2{ { a }_{ 2 } }^{ 2 }-4{ { a }_{ 1 } }^{ 2 }+6{ a }_{ 1 }{ a }_{ 2 } \right) 63 \\ +\left( { { a }_{ 2 } }^{ 2 }+4{ { a }_{ 1 } }^{ 2 }-4{ a }_{ 1 }{ a }_{ 2 } \right) 9=990\\ =(1+501+9-126){ { a }_{ 2 } }^{ 2 }+{ { a }_{ 1 } }^{ 2 } (1+501-252+36)+2{ a }_{ 1 }{ a }_{ 2 }(-501+189-18)=990\quad \left( \because { a }_{ 1 }=15 \right) \\ =385{ { a }_{ 2 } }^{ 2 }+64350-9900{ a }_{ 2 }=990\\ =385{ { a }_{ 2 } }^{ 2 }-9900{ a }_{ 2 }+63360=0\\ { a }_{ 2 }=\cfrac { +9900\pm \sqrt { { (9900) }^{ 2 }-4\times 385\times 63360 } }{ 2\times 385 } \\ { a }_{ 2 }=12\\ \cfrac { \left( { a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+....+{ a }_{ 11 } \right) }{ 11 } \\ =\cfrac { \left[ { a }_{ 1 }+{ a }_{ 2 }+\overset { 11 }{ \underset { k=3 }{ \sum   } } (k-1){ a }_{ 2 }-(k-2){ a }_{ 1 } \right] }{ 11 } \\ =\cfrac { \left[ { a }_{ 1 }+{ a }_{ 2 }+({ a }_{ 2 }-{ a }_{ 1 })\overset { 11 }{ \underset { k=3 }{ \sum   } } k+(2{ a }_{ 1 }-{ a }_{ 2 })\overset { 11 }{ \underset { k=3 }{ \sum   } } (1) \right] }{ 11 } \\ =\cfrac { 15+12+(-3){ \left[ \cfrac { k(k+1) }{ 2 } \right] }_{ 2 }^{ 11 }+(2\times 15-12){ \left[ k \right] }_{ 2 }^{ 11 } }{ 11 } \\ =\cfrac { (27-3\times 63+18\times 9) }{ 11 } \\ =0$$
Question 3
1 / -0

It is known that $$\sum_{r=1}^{\infty }\frac{1}{\left ( 2r-1 \right )^{2}}=\frac{\pi ^{2}}{8}$$. Then $$\sum_{r=1}^{\infty }\frac{1}{r^{2}}$$ is equal to

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

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

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

D
none of these

Solution

Given
$$ \dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+... = \dfrac{\pi^{2}}{8} $$

Now, $$ S_{n} = \dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+... $$

$$\Rightarrow S_{n} = \left\{\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+... \right\} + \left\{\dfrac{1}{2^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{6^{2}}+... \right\} $$

$$\Rightarrow S_{n} = \dfrac{\pi^{2}}{8} + \dfrac{1}{2^{2}}\left\{\dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+... \right\} $$

$$ S_{n} = \dfrac{\pi^{2}}{8}+\dfrac{1}{4}S_{n} $$

$$ \Rightarrow \dfrac{3}{4}S_{n}= \dfrac{\pi^{2}}{8} $$

$$ \therefore S_{n} = \dfrac{\pi^{2}}{6} $$
Question 4
1 / -0

Evaluate $$\sum_{r=1}^{n}\left [ \sum_{k=1}^{r}k \right ] \left [ \log_{1/2}\sqrt{(4x-x^{2})} \right ]^{r}$$. Find $$x$$ for which summation is a finite number as $$n\rightarrow \infty $$

A
$$x\in \left ( 0,2+\displaystyle \frac{\sqrt{15}}{2} \right )$$

B
$$x\in \left ( 0,2-\displaystyle \frac{\sqrt{15}}{2} \right )$$

C
$$x\in \left ( 0,-2+\displaystyle \frac{\sqrt{15}}{2} \right )$$

D
$$x\in \left ( 0,-2-\displaystyle \frac{\sqrt{15}}{2} \right )$$

Solution

We have,
$$

\sum_{r=1}^{n}\left [ \sum_{k=1}^{r}k \right ] \left [

\log_{1/2}\sqrt{(4x-x^{2})} \right ]^{r}$$
$$=\sum_{r=1}^{n} \displaystyle \frac{r(r+1)}{2}\left [ \log_{1/2} \sqrt{(4x-x^{2})}\right ]^{r}$$
$$=\sum_{r=1}^{n} \displaystyle \frac{r(r+1)}{2}A^{r}$$ [where $$A= \log_{1/2} \sqrt{(4x-x^{2})}$$]
$$=\displaystyle \frac{1}{2}\sum_{r=1}^{n}r(r+1)A^{r}$$
$$=\displaystyle \frac{1}{2}\left [ (1.2)A+(2.3)A^{2}+(3.4)A^{3}+....+n(n+1)A^{n} \right ]$$
$$S=\displaystyle \frac{1}{2}S_{1}$$ (say) $$\cdots (1)$$
$$\therefore S_{1}=(1.2)A+(2.3)A^{2}+(3.4)A^{3}+....+(n-1)nA^{n-1}+n(n+1)A^{n}$$
$$\Rightarrow S_{1}=2A+6A^{2}+12A^{3}+.....+(n-1)nA^{n-1}+n(n+1)A^{n}$$
$$\therefore AS_{1}=0+2A^{2}+6A^{3}+....+(n-1)nA^{n}+n(n+1)A^{n+1}$$
Subtracting, we get,
$$(1-A)S_{1}=2A+4A^{2}+6A^{3}+....+2nA^{n}-n(n+1)A^{n+1}$$
$$\Rightarrow (1-A)S_{1}=2[A+2A^{2}+3A^{3}+...+nA^{n}]-n(n+1)A^{n+1}$$
$$=2S_{2}-n(n+1)A^{n+1}$$ (say) $$\cdots (2)$$
$$\therefore S_{2}=A+2A^{2}+3A^{3}+....+(n-1)A^{n-1}+n A^{n}$$
from $$(2)$$,
$$(1-A)S_{1}=\displaystyle \frac{2A(1-A^{n})}{(1-A)^{2}}-\displaystyle \frac{2nA^{n+1}}{(1-A)}-n(n+1)A^{n+1}$$
$$\therefore

S_{1}=\displaystyle \frac{2A(1-A^{n})}{(1-A)^{3}}-\displaystyle

\frac{2nA^{n+1}}{(1-A)^{2}}-\displaystyle \frac{n(n+1)A^{n+1}}{(1-A)}$$
now from $$(1)$$,
$$S=\displaystyle \frac{1}{2}S_{1}$$
$$\therefore

S=\displaystyle \frac{A(1-A^{n})}{(1-A)^{3}}-\displaystyle

\frac{nA^{n+1}}{(1-A)^{2}}-\displaystyle \frac{n(n+1)A^{n+1}}{2(1-A)}$$
Now, for $$S$$ to be finite we must have $$|A| < 1$$
$$S_{\infty }=\displaystyle \frac{A(1-0)}{(1-A)^{3}}-0-0$$
$$=\displaystyle

\frac{A}{(1-A)^{3}}=\displaystyle

\frac{\log_{1/2}\sqrt{(4x-x^{2})}}{(1-\log_{1/2}\sqrt{(4x-x^{2})})^{3}}=finite$$

$$(\because A<1)$$
i.e $$\log_{1/2}\sqrt{(4x-x^{2})} < 1$$
$$\Rightarrow \sqrt{4x-x^{2}} > \displaystyle \frac{1}{2}$$
$$\Rightarrow 4x-x^{2} > \displaystyle \frac{1}{4}$$
$$\Rightarrow 4x^{2}-16x+1 < 0$$
$$\therefore 2-\displaystyle \frac{\sqrt{15}}{2} < x < 2+\displaystyle \frac{\sqrt{15}}{2} \cdots (3)$$
and $$4x-x^{2} > 0$$
$$\Rightarrow x^{2}-4x < 0$$
$$\Rightarrow x(x-4) < 0$$
$$\therefore 0<x<4 \cdots (4)$$
combining $$(3)$$ and $$(4)$$ we get
$$x\in \left ( 0,2+\displaystyle \frac{\sqrt{15}}{2} \right )$$
Question 5
1 / -0

if $$\displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{2^2}\, +\, \displaystyle \frac{1}{3^2}$$ + .......... upto $$\infty\, =\, \displaystyle \frac{\pi^2}{6}$$, then $$\displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{3^2}\, +\, \displaystyle \frac{1}{5^2}$$ + ........... = .......... .

A
$$\pi^2 / 8$$

B
$$\pi^2 / 12$$

C
$$\pi^2 / 3$$

D
$$\pi^2 / 9$$

Solution

$$\displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{2^2}\, +\, \displaystyle \frac{1}{3^2}\, +\, \displaystyle \frac{1}{4^2}\, ..........\, \infty\, =\, \displaystyle \frac{\pi^2}{6}$$
Now $$ \displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{3^2}\, +\, \displaystyle \frac{1}{5^2}\, ...........$$
$$=\, \displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{2^2}\, +\,\displaystyle \frac{1}{3^2}\,+\, \displaystyle \frac{1}{4^2}\, ..........\,\infty\, -\, \left ( \displaystyle \frac{1}{2^2}\, +\, \displaystyle \frac{1}{4^2}\, +\, \displaystyle \frac{1}{6^2}\, ..........\, \infty \right )$$
$$=\, \displaystyle \frac{\pi^2}{6}\, -\, \displaystyle \frac{1}{2^2}\, \left ( \displaystyle \frac{1}{1^2}\, +\, \displaystyle \frac{1}{2^2}\, +\, \displaystyle \frac{1}{3^2} \, ..........\, \infty \right )$$
$$=\, \displaystyle \frac{\pi^2}{6}\, -\, \displaystyle \frac{1}{4}\, .\, \displaystyle \frac{\pi^2}{6}\, =\, \displaystyle \frac{\pi^2}{8}$$
Question 6
1 / -0

Let $$\displaystyle \left \{ a_{n} \right \}\: and\: \left \{ b_{n} \right \}$$ are two sequences given by $$\displaystyle a_{n}=\left ( x \right )^{1/2^{n}}+\left ( y \right )^{1/2^{n}}\: \: and\: \: b_{n}=\left ( x \right )^{1/2^{n}}-\left ( y \right )^{1/2^{n}}$$ for all n $$\displaystyle \epsilon $$ N. The value of $$\displaystyle a_{1}a_{2}a_{3}...a_{n}$$ is equal to

A
x - y

B
$$\displaystyle \frac{x+y}{b_{n}}$$

C
$$\displaystyle \frac{x-y}{b_{n}}$$

D
$$\displaystyle \frac{xy}{b_{n}}$$

Solution

$${ a }_{ n }=\cfrac { \left( { x }^{ 1/{ 2 }^{ n } }+{ y }^{ 1/{ 2 }^{ n } } \right) \left( { x }^{ 1/{ 2 }^{ n } }-{ y }^{ 1/{ 2 }^{ n } } \right) }{ \left( { x }^{ 1/{ 2 }^{ n } }-{ y }^{ 1/{ 2 }^{ n } } \right) } \\ { a }_{ n }=\cfrac { \left( { x }^{ 1/{ 2 }^{ n-1 } }-{ y }^{ 1/{ 2 }^{ n-1 } } \right) }{ \left( { x }^{ 1/{ 2 }^{ n } }-{ y }^{ 1/{ 2 }^{ n } } \right) } =\cfrac { { b }_{ n-1 } }{ { b }_{ n } } \\ { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }......{ a }_{ n }=\cfrac { { b }_{ 0 } }{ { b }_{ 1 } } \times \cfrac { { b }_{ 1 } }{ { b }_{ 2 } } \times \cfrac { { b }_{ 2 } }{ { b }_{ 3 } } \times .....\times \cfrac { { b }_{ n-1 } }{ { b }_{ n } } \\ { a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }......{ a }_{ n }=\cfrac { { b }_{ 0 } }{ { b }_{ n } } =\cfrac { (x-y) }{ { b }_{ n } } $$
Question 7
1 / -0

If a number sequence begins $$1, 3, 4, 6, 7, 9, 10, 12 . . .$$, which of the following numbers does NOT appear in the sequence?

A
$$34$$

B
$$43$$

C
$$57$$

D
$$65$$

E
$$72$$

Solution

$$1,3,4,6,7,9,10,12...$$
$$=\left( 1,4,7,10... \right) ,\left( 3,6,9,12... \right) $$
If a number can be written in the form of $$\left( 3k \right) $$ or $$\left( 3k+1 \right) $$., then it will appear in the sequence.
Now check options
For $$34=3\left( 11 \right) +1=3k+1$$, where $$k=11$$
For $$43=3\left( 14 \right) +1=3k+1$$, where $$k=14$$
For $$57=3\left( 19 \right) =3k$$, where $$k=19$$
For $$72=3\left( 24 \right) $$, where $$k=24$$
For Here $$65=3\left( 21 \right) +2=3k+2$$
$$\left( 3k+2 \right) $$ will not appear.
Therefore, $$ 65$$ will not appear in the sequence.
Question 8
1 / -0

Sum of the series $$\displaystyle \sum_{r=1}^{88}\left ( -1 \right )^{r+1}\frac{1}{\sin ^{2}\left ( r+1 \right )^{\circ}-\sin ^{2}1^{\circ}}$$ is equal to

A
$$\displaystyle \frac{\cot 2^{\circ}}{\sin 2^{\circ}}$$

B
$$\displaystyle \frac{-\cot 2^{\circ}}{\sin 2^{\circ}}$$

C
$$\displaystyle \cot 2^{\circ}$$

D
$$\displaystyle \frac{\cot 2^{\circ}}{\sin ^{2}2^{\circ}}$$

Solution

$$ \displaystyle \sum_{88}^{r=1}(-1)^{r+1}\frac{1}{\sin ^{2}(r+1)^{\circ}-\sin ^{2}1^{\circ}} $$
$$

\displaystyle =\left (\frac{1}{\sin 1^{\circ}\sin

3^{\circ}}+\frac{1}{\sin 3^{\circ}\sin 5^{\circ}}+...+\frac{1}{\sin

87^{\circ}\sin 89^{\circ}} \right )-\left (\frac{1}{\sin 2^{\circ}\sin

4^{\circ}}+\frac{1}{\sin 4^{\circ}\sin 6^{\circ}}+.....+\frac{1}{\sin

88^{\circ}\sin 90^{\circ}} \right ) $$
$$ \displaystyle =

\frac{1}{\sin 2^{\circ}}\left [ \left ( \cot 1^{\circ}-\cot 3^{\circ}

\right ) +\left ( \cot 3^{\circ}-\cot 5^{\circ} \right )+....+\left (

\cot 87^{\circ}-\cot 89^{\circ} \right ) \right ]$$
$$ \displaystyle

=\left [ \left ( \cot 2^{\circ}-\cot 4^{\circ} \right )+ \left ( \cot

4^{\circ}-\cot 6^{\circ} \right )+........+\left ( \cot 88^{\circ}-\cot

90^{\circ} \right ) \right ]$$
$$ \displaystyle =\frac{\cot 2^{\circ}-0}{\sin 2^{\circ}}=\frac{\cot 2^{\circ}}{\sin 2^{\circ}} $$
Question 9
1 / -0

The expression
$$\displaystyle \frac {2^2 + 1} {2^2 - 1} + \frac {3^2 + 1} {3^2 - 1} + \frac {4^2 + 1} {4^2 - 1} + ........... + \frac {(2011)^2 + 1} {(2011)^2 - 1} $$
lies in the interval

A
$$\displaystyle (2011, 2010 \frac {1} {2} )$$

B
$$\displaystyle \left( 2011 - \frac {1} {2011} , 2011 - \frac {1} {2012} \right)$$

C
$$\displaystyle (2011, 2011 \frac {1} {2} )$$

D
$$\displaystyle (2012, 2012 \frac {1} {2} )$$

Solution

We can write the expressions as

$$\displaystyle \sum _{r = 2}^{2011} {\frac {r^2 + 1}{r^2 - 1}} = \sum \left[1 + \frac {2} {(r + 1) (r - 1)} \right] $$

$$\displaystyle = \sum \left[ 1 + \frac {1} {r-1} - \frac {1} {r+1} \right] $$

Putting $$ r = 2, 3, ............, 2011$$

$$\displaystyle = 2010 + 1 + \frac {1} {2} - \frac {1} {2012} - \frac {1} {2011} $$

$$\displaystyle = 2011 + \frac {1} {2} - \left[ \frac {1} {2012} + \frac {1} {2011} \right] $$

this lies between $$\displaystyle (2011, 2011 \frac {1} {2}) $$
Question 10
1 / -0

The value of $$1000\left[\dfrac {1}{1\times 2}+\dfrac {1}{2\times 3}+\dfrac {1}{3\times 4}+...+\dfrac {1}{999\times 1000}\right]$$ is equal to

A
$$1000$$

B
$$999$$

C
$$1001$$

D
$$\dfrac{1}{999}$$

Solution

$$=1000[\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{999\times1000}]$$

$$=1000[\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+...+\dfrac{1000-999}{999\times1000}]$$

$$=1000[\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}]$$,

$$=1000[1-\dfrac{1}{1000}]$$ Only end terms remain, as all intermediate terms cancel out.

$$=1000.\dfrac{999}{1000}$$

$$=999$$

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Sequences and Series Test - 53

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Sequences and Series Test - 53

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SHARING IS CARING

Question 1

AP

GP

HP

none of these

Solution

Question 2

1

5

9

0

Solution

Question 3

$$\dfrac{\pi ^{2}}{24}$$

$$\dfrac{\pi ^{2}}{3}$$

$$\dfrac{\pi ^{2}}{6}$$

none of these

Solution

Question 4

$$x\in \left ( 0,2+\displaystyle \frac{\sqrt{15}}{2} \right )$$

$$x\in \left ( 0,2-\displaystyle \frac{\sqrt{15}}{2} \right )$$

$$x\in \left ( 0,-2+\displaystyle \frac{\sqrt{15}}{2} \right )$$

$$x\in \left ( 0,-2-\displaystyle \frac{\sqrt{15}}{2} \right )$$

Solution

Question 5

$$\pi^2 / 8$$

$$\pi^2 / 12$$

$$\pi^2 / 3$$

$$\pi^2 / 9$$

Solution

Question 6

x - y

$$\displaystyle \frac{x+y}{b_{n}}$$

$$\displaystyle \frac{x-y}{b_{n}}$$

$$\displaystyle \frac{xy}{b_{n}}$$

Solution

Question 7

$$34$$

$$43$$

$$57$$

$$65$$

$$72$$

Solution

Question 8

$$\displaystyle \frac{\cot 2^{\circ}}{\sin 2^{\circ}}$$

$$\displaystyle \frac{-\cot 2^{\circ}}{\sin 2^{\circ}}$$

$$\displaystyle \cot 2^{\circ}$$

$$\displaystyle \frac{\cot 2^{\circ}}{\sin ^{2}2^{\circ}}$$

Solution

Question 9

$$\displaystyle (2011, 2010 \frac {1} {2} )$$

$$\displaystyle \left( 2011 - \frac {1} {2011} , 2011 - \frac {1} {2012} \right)$$

$$\displaystyle (2011, 2011 \frac {1} {2} )$$

$$\displaystyle (2012, 2012 \frac {1} {2} )$$

Solution

Question 10

$$1000$$

$$999$$

$$1001$$

$$\dfrac{1}{999}$$

Solution

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