Sequences and Series Test

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Sequences and Series Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

The sum of the series $$\dfrac{1}{(1\times 2)}+\dfrac{1}{(2\times 3)}+\dfrac{1}{(3\times 4)}+.......+\dfrac{1}{(100\times 101)}$$ is equal to

A
$$\dfrac{200}{101}$$

B
$$\dfrac{100}{101}$$

C
$$\dfrac{50}{101}$$

D
$$\dfrac{25}{101}$$

Solution

$$\dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\cdots +\dfrac{1}{100\times 101}=\dfrac{2-1}{1\times 2}+\dfrac{3-2}{2\times 3}+\cdots+\dfrac{101-100}{100\times 101}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots +\dfrac{1}{100}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}$$
Question 2
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If $$(1+3+5+....+p)+(1+3+5+....+q)=(1+3+5+.....+r),$$ where each set of parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of $$p+q+r$$, (where $$P>6$$ ) is :

A
$$12$$

B
$$21$$

C
$$27$$

D
$$24$$

Solution

$$ 1+3+5+...+2n-1 = n^{2} $$
no. of terms $$ = \dfrac{n+1}{2} $$
According to question we can write
$$\left (\dfrac{p+1}{2}\right)^{2}+\left(\dfrac{q+1}{2}\right)^{2}\pm \left(\dfrac{r+1}{2}\right)^{2} $$
It forms Pythagoras triplet Thus for min value
take 3, 4 and 5 as our Pythagoras triplet
$$ \therefore \dfrac{p+1}{2} = 4 $$ $$ \dfrac{q+1}{2} = 3 $$ $$ \dfrac{r+1}{2} = 5 $$
$$ p = 7 $$ $$ q = 5 $$ $$ r = 9 $$
$$ p + q + r = 21 $$
Question 3
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Let $$T_{n}=\displaystyle \sum _{ r=1 }^{ n }{ \dfrac { n }{ { r }^{ 2 }-2r.n+2{ n }^{ 2 } } ,{ S }_{ n }=\displaystyle \sum _{ r=0 }^{ n-1 }{ \dfrac { n }{ { r }^{ 2 }-2r.n+{ 2n }^{ 2 } } } }$$, then

A
$$T_{n} > S_{n} \forall n \epsilon N$$

B
$$T_{n} > \dfrac{\pi}{4}$$

C
$$S_{n} < \dfrac{\pi}{4}$$

D
$$\displaystyle \lim _{ n-\infty }{ { S }_{ n }=\dfrac { \pi }{ 4 } }$$

Solution

Sol $$T_{n}=\displaystyle \sum _{ r=1 }^{ n }{ \dfrac { n }{ { r }^{ 2 }-2r.n+2{ n }^{ 2 } } ,{ S }_{ n }=\displaystyle \sum _{ r=0 }^{ n-1 }{ \dfrac { n }{ { r }^{ 2 }-2r.n+{ 2n }^{ 2 } } } }$$
On expanding $$ T_{n} $$ and $$ S_{2} $$ we get
$$ T_{n} = \dfrac{n}{2n^{2}-2n+1}+\dfrac{n}{2n^{2}-4n+4}+\dfrac{n}{2n^{2}-6n+9}+...+\dfrac{n}{n^{2}} ... (1) $$
and
$$ S_{n} = \dfrac{n}{2n^{2}}+\dfrac{n}{2n^{2}-2n+1}+\dfrac{n}{2n^{2}-4n+4}+...+\dfrac{n}{2n^{1}-2n^{2}+2n+(n-1)^{2}} $$
$$ S_{n} = \dfrac{n}{2n^{2}}+\dfrac{n}{2n^{2}-2n+1}+\dfrac{n}{2n^{2}-2n+4}+...+\dfrac{n}{n^{2}+1}...(2) $$
from eq (1)
on comparing $$ T_{n} $$ & $$ S_{2} $$ we can clearly
see that
$$ T_{n}\Rightarrow S_{n} $$
as $$ T_{n} = P+\dfrac{1}{n} $$
and $$ S_{n} = P+\dfrac{1}{2n} $$
as $$ P = $$ common part in eq (1) & (2)
to $$ [T_{n}> S_{n} \forall\ n \epsilon N] $$
Hence from equation (1) & (2)
$$ [T_{n} = S_{n}+\dfrac{1}{2n}] $$
Question 4
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Let $${n}^{th}$$ term of sequence $$1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,.....$$ is given by $${t}_{n}$$ , then ?

A
$${t}_{100}=14$$

B
$${t}_{200}=20$$

C
$${t}_{300}=24$$

D
$${t}_{400}=28$$

Solution

$$no.\>of\>terms\>with\>same\>values\>are\>in\>AP(let's\>call\>them\>as\>a\>group)\\100=(\frac{n(n+1)}{2})\\n(n+1)=200\\for\>n\>=14\\n(n+1)=210\>>\>200\\hence\>all\>terms\>with\>values\>14,\>will\>cover\>group\>including\>100th\>term$$
Question 5
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If $$S_{n}=\dfrac {3}{2}+\dfrac {33}{2^{2}}+\dfrac {333}{2^{2}}+....$$ upto $$n$$ terms $$=\dfrac {a^{n+1}+b^{b-n}-c}{d}$$ (where $$a,b,c,d \epsilon\ N$$), then ?

A
$$a+b+c+d=28$$

B
$$b+d=a+c$$

C
$$a-c=b+d$$

D
$$d-c=a-b$$

Solution
Question 6
1 / -0

Sum of the series $$1+2.2+3.2^{2}+4.2^{3}+..+100.2^{99}$$ is

A
$$100.2^{100}+1$$

B
$$99.2^{100}+1$$

C
$$99.2^{100}-1$$

D
$$100.2^{100}-1$$

Solution

It is been clear that the series is in Arithmetic-Geometric progression with
$$A.P=\quad 1+2+3+.......+100\\ and,\quad G.P=\quad 1+2+{ 2 }^{ 2 }+.....+{ 2 }^{ 99 }\\ Let,\quad S=1+2.2+3.{ 2 }^{ 2 }+4.{ 2 }^{ 3 }+.....+100.{ 2 }^{ 99 }\quad \quad \quad \quad \quad \longrightarrow \quad (1)\\ \text{Multiply the above eauation by} 2\\ 2S=2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+.....+99.{ 2 }^{ 99 }+100.{ 2 }^{ 100 }\quad \quad \quad \quad \quad \quad \longrightarrow \quad (2)\\ now\quad subtract\quad (2)\quad from\quad (1)\quad we\quad get,\\ \quad \quad \quad \quad S-2S=(1+2+{ 2 }^{ 2 }+....+{ 2 }^{ 99 })-100.{ 2 }^{ 100 }\\ \Rightarrow \quad -S=\frac { ({ 2 }^{ 100 }-1) }{ 2-1 } -100.{ 2 }^{ 100 }\quad \quad (since\quad the\quad above\quad series\quad in\quad bracket\quad is\quad in\quad G.P\quad so\quad { S }_{ n }=\frac { a({ r }^{ n }-1) }{ r-1 } \quad and\quad this\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad is\quad a\quad G.P\quad with\quad 100\quad terms\quad and\quad a=1;\quad r=2)\\ \Rightarrow \quad S=\quad -{ 2 }^{ 100 }+1+100.{ 2 }^{ 100 }\\ \therefore \quad S=\quad 99.{ 2 }^{ 100 }+1\\$$
Question 7
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The value of sum $$\sum _{ n=1 }^{n=\infty }{ \frac { 2 }{ { 3 }^{ n } } is\ equal\ to: } $$

A
1

B
3

C
15/4

D
7/3

Solution
Question 8
1 / -0

Pam likes the numbers 1689 and 6891. Knowing this, which pair of numbers will she like among the ones below?

A
1981 and 1891

B
19 and 91

C
190 and 160

D
1198911 and 1168611

Solution
Question 9
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The sum of the first $$n$$ terms of the series $$\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{7}{8}+\dfrac{15}{16}+....$$ is

A
$$2^n-n-1$$

B
$$1-2^{-n}$$

C
$$n+2^{-n}-1$$

D
$$2^n-1$$
Question 10
1 / -0

Find sum $${1}^{2}+{3}^{2}+{5}^{2}+...+{(2n-1)}^{2}$$

A
$$\dfrac{n(2n+1)(2n+1)}{3}$$.

B
$$\dfrac{n(2n-1)(2n+1)}{3}$$.

C
$$\dfrac{2n(2n-1)(2n+1)}{3}$$.

D
None of these

Solution

$${1^2} + {2^2} + {3^2} + {4^2} + {5^2} + ... + {\left( {2n - 1} \right)^2}$$
Here $${a_n} = {\left( {2n - 1} \right)^2}$$
Sum of the given series $$=$$$$\sum\limits_{n = 1}^n {{a_n}} $$
$$ = \sum\limits_{n = 1}^n {{{\left( {2n - 1} \right)}^2}} $$
$$ = \sum\limits_{n = 1}^n {\left( {4{n^2} + 1 - 4n} \right)} $$
$$= 4\sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n 1 - 4\sum\limits_{n = 1}^n n $$
$$ = \dfrac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + n - \dfrac{{4n\left( {n + 1} \right)}}{2}$$
$$ = \dfrac{n}{6}\left[ {8{n^2} + 12n + 4 + 6 - 12n - 12} \right]$$
$$ = \dfrac{n}{6}\left[ {8{n^2} - 2} \right]$$
$$ = \dfrac{n}{3}\left[ {4{n^2} - 1} \right]$$
$$= \dfrac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}$$