Sequences and Series Test

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Sequences and Series Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \sum_{r = 0}^{n}{ \left( \frac{ 2^{r-2} . ^nC_r}{(r+1)(r+2)} \right) }$$ is equal to

A
$$\displaystyle \frac{3^{n+2} - 2n + 5}{(n+1)(n+2)}$$

B
$$\displaystyle \frac{3^{n+2} - 4n + 5}{(n+1)(n+2)}$$

C
$$\displaystyle \frac{3^{n+2} - 2n - 5}{(n+1)(n+2)}$$

D
None of these
Question 2
1 / -0

If the $$p^{th}$$ term of the series of positive numbers $$25, 22\dfrac {3}{5}, 20\dfrac {1}{2}, 18\dfrac {1}{4}$$, .... is numerically the smallest, then the $$p^{th}$$ is.

A
$$\dfrac{1}{4}$$

B
$$\dfrac{1}{7}$$

C
$$\dfrac{1}{3}$$

D
$$\dfrac{1}{5}$$

Solution

$$d = 22\dfrac{3}{4} - 25$$
$$=-2\dfrac{1}{4} = -\dfrac{9}{4}$$
$$t_n \ge 0$$
$$\therefore t_n = 25 + (n-n)\left(-\dfrac{9}{4}\right)$$
$$25-\dfrac{9}{4} (n-1) \ge 0$$
$$\Rightarrow 25\ge \dfrac{9}{4} (p-1) \ge 0$$
$$\Rightarrow \dfrac{100}{9} \ge p -1$$
$$\Rightarrow \dfrac{109}{9} \ge p$$
$$12 \dfrac{1}{9} \ge p$$
12th term is positive.
$$25 + (12 - 1) \times \left(-\dfrac{9}{4}\right) = \dfrac{1}{4}$$
Question 3
1 / -0

13, 74, 290, 650,.......

A
$$1248$$

B
$$1370$$

C
$$1346$$

D
$$1452$$

E
$$1625$$

Solution

Here we observe the first term 13 can be written as: $$2^2+3^2$$
Similarly other numbers can be written as:
$$74: 5^2+7^2$$
$$290: 11^2+13^2$$
$$650: 17^2+19^2$$
Here, we observe that all above term of series can be written as sum of squares of two prime numbers which comes next to it's previous pair.
Like: $$(2,3);(5,7);(11,13);(17,19)$$ So, other pair of prime number will be $$ (23,29).$$
Hence next term of given series will be:
$$\Rightarrow 23^2+29^2=1370$$
So, correct answer is $$1370$$.
Question 4
1 / -0

If $$\begin{vmatrix} x \end{vmatrix}<1$$ then the coefficient of $$x^5$$ in the expansion of $$\dfrac{3x}{(x-2) (x-1)}$$ is

A
$$\dfrac{33}{32}$$

B
$$-\dfrac{33}{32}$$

C
$$\dfrac{31}{32}$$

D
$$-\dfrac{33}{34}$$

Solution

$$\dfrac{3x}{(x-2) (x+1)}= \dfrac {A}{x-2}+\dfrac{B}{x+1}$$

$$A=\dfrac{3(2)}{2+1}=2; B=\dfrac{3(-1)}{-1-2}=\dfrac{-3}{-3}=1$$

$$\dfrac{3x}{(x-2)(x+1)}=\dfrac{2}{x-2}+\dfrac{1}{x+1}$$

$$=\dfrac{2}{2(1-\dfrac {x }{ 2 } )} +\dfrac {1}{x+1} = -\begin{pmatrix} 1-\dfrac { x }{ 2 } \end{pmatrix}^{-1} +(1+x)^{-1}$$

$$=-\begin{pmatrix} 1+\dfrac{x}{2}+\begin{pmatrix} \dfrac {x}{2}\end{pmatrix}^2+......+\begin{pmatrix} \dfrac{x}{2} \end{pmatrix}^5+...... \end{pmatrix}+(1-x+x^2-x^3+x^4-x^5+......)$$

Coefficient of $$x^5=\dfrac{-1}{32}-1=\dfrac{-33}{32}$$
Question 5
1 / -0

If the coefficients of $$x^9,x^{10},x^{11}$$ in the expansion of $$(1+x)^n $$ are in arithmetic progression then $$n^2-41n=$$

A
$$398$$

B
$$298$$

C
$$-398$$

D
$$198$$

Solution

$$^nC_9,^nC_{10},^nC_{11}$$ are in A.P

$$\Rightarrow 2\ ^{n}C_{10}=^nC_9+^nC_{11} \Rightarrow=\dfrac{^nC_9}{^nC_{10}}+\dfrac{^nC_{11}}{^nC_{10}}$$

$$\Rightarrow 2=\dfrac{10}{n-9}+\dfrac{n-10}{11} \Rightarrow 2=\dfrac{110+n^2-19n+90}{11n-99}$$

$$\Rightarrow n^2-41n=-398$$
Question 6
1 / -0

If $$\sin^{-1}\begin{pmatrix} x-\frac{x^2}{2}+\frac{x^3}{4}-..........\infty \end{pmatrix}+\cos^{-1}\begin{pmatrix} x^2-\frac{x^4}{2}+\frac{x^6}{4}........\infty\end{pmatrix} =\frac{\pi}{2}$$ and $$0<x<\sqrt{2}$$ then x=

A
$$\dfrac{1}{2}$$

B
$$1$$

C
$$\dfrac{-1}{2}$$

D
$$-1$$

Solution
Question 7
1 / -0

The value of $$\cfrac { 1 }{ \left( 2n-1 \right) !0! } +\cfrac { 1 }{ \left( 2n-3 \right) !2! } +\cfrac { 1 }{ \left( 2n-5 \right) !4! } +....+\cfrac { 1 }{ 1!\left( 2n-2 \right) ! } $$ equal to

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

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

C
$${ 2 }^{ 2n-3 }$$

D
$$\cfrac { { 2 }^{ 2n-2 } }{ \left( 2n-1 \right) ! } $$

Solution

We know that,
$$^nC_0+^nC_1+^nC_2..........^nC_n=2^n$$

$$^nC_0+^nC_2+^nC_4..........^nC_{n-1}=$$$$^nC_1+^nC_3+^nC_5..........^nC_n=2^{n-1}$$

Using the above relation,

$$^{2n-1}C_0+^{2n-1}C_1+^{2n-1}C_2..........^{2n-1}C_n=2^{(2n-1)-1}$$

Expanding the coefficients we get the desired result,

$$\dfrac{(2n-1)!}{(2n-1)!0!}+$$$$\dfrac{(2n-1)!}{(2n-3)!2!}+$$$$\dfrac{(2n-1)!}{(2n-5)!4!}+$$$$...............+\dfrac{(2n-1)!}{1!(2n-2!)}=2^{2n-2}$$

$$\dfrac{1}{(2n-1)!0!}+$$$$\dfrac{1}{(2n-3)!2!}+$$$$\dfrac{1}{(2n-5)!4!}+$$$$...............+\dfrac{1}{1!(2n-2!)}=\dfrac{2^{2n-2}}{(2n-1)!}$$
Question 8
1 / -0

The sum of first $$20$$ terms of the series $$1,6,13,22$$- is

A
$$5580$$

B
$$5780$$

C
$$7789$$

D
$$1237$$

Solution

Given series, $$ S = 1,6,13,22,...$$
Sum of series, $$ S_{n} = 1+6+13+22+...+ T_{n}$$
$$ S_{n} = 1+6+13+...+T_{n}+T_n$$
$$ 0 = 1+5+7+9+,,,T_{n}$$
$$ T_{n} = 1+ [5+7+9+...(n-1)terms]$$
(sum of AP) $$ = 1+\left [ \left ( \dfrac{n-1}{2} \right )(2(5) + (n-2)2) \right ]$$
$$ = 1+ \left [ (n-1)(n-3) \right ]$$
$$ = (n+1)^2 -3 $$
sum of 20 terms in series $$ = \displaystyle \sum_{n=1}^{20} (n^2 + 2n-2)$$
$$ = \displaystyle \sum_{n=1}^{20}n^2 +2\sum_{n=1}^{20} n - 2 \times 20 $$
$$ = 5780 $$
Question 9
1 / -0

$$2.4+4.7+6.10+.....$$ upto $$(n-1)$$ terms

A
$$2{ n }^{ 3 }+2{ n }^{ 2 }$$

B
$$\cfrac { 1 }{ 6 } \left( { n }^{ 3 }+3{ n }^{ 2 }+1 \right) $$

C
$$2{ n }^{ 3 }-2{ n }^{ 2 }$$

D
$$\cfrac { 1 }{ 6 } \left( { n }^{ 2 }-3{ n }+1 \right) $$

Solution

The general term of the series is $$T_n=2n(3n+1)=6n^2+2n$$
The sum of series to $$n$$ terms is $$6\times\dfrac{n(n+1)(2n+1)}6+2\times\dfrac{n(n+1)}2=n(n+1)[(2n+1)+1]=2n(n+1)^2$$
Now, we need the sum till $$n-1$$ terms, so we substitute $$n-1$$ in place of $$n$$
Hence, sum of series is $$2(n-1)n^2=2n^3-2n^2$$
Question 10
1 / -0

Sum of the series $$\sum _{ r=1 }^{ n }{ \left( { r }^{ 2 }+1 \right) r! } $$ is ______

A
$$(n+1)!$$

B
$$(n+2)!-1$$

C
$$n(n+1)!$$

D
none of these

Solution

$$\sum _{ r=1 }^{ n }{ \left( { r }^{ 2 }+1 \right) r! } $$

$$\Rightarrow \sum_{ r=1 }^{ n }{ ( { r }^{ 2 }+1+2r-2r ) r! }$$

$$\Rightarrow \sum_{ r=1 }^{ n }{ ( { ( r+1) }^{ 2 }.r!-2r.r! ) }$$

$$\Rightarrow \sum_{ r=1 }^{ n }{ ( { ( r+1) }^{ }.(r+1)!-2r.r! ) }$$

$$\Rightarrow \sum_{ r=1 }^{ n }{ [ ( r+2-1 )( r+1 )!-2( ( r+1-1 )r! ) ] }$$

$$\Rightarrow \sum_{ r=1 }^{ n }{ [ ( r+2 )( r+1 )!-( r+1 )!-2( ( r+1 )r!-r! ) ] }$$

$$\Rightarrow \sum_{ r=1 }^{ n }{ \left [ ( r+2 )!-( r+1 )!-2.( r+1 )!+2.r!) \right] }$$

$$T_1=3!-2!- 2.2!+2.1!$$

$$T_2=4!-3!-2.3!+2.2!$$

$$T_3=5!-4!-2.4!+2.3!$$

.

.
.
.
.
.

.

$$T_n=(n+2)!-(n+1)!-2.(n+1)!+2.n!$$

$$-------------------$$

$$Sum=S_n=(n+2)!-2!-2(n+1)!+2.1!$$

$$S_n=(n+2)!-2(n+1)!-2+2$$

$$S_n=(n+1)![n+2-2]$$

$$S_n=n.(n+1)!$$

Correct answer is C

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Sequences and Series Test - 31

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Sequences and Series Test - 31

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Question 1

$$\displaystyle \frac{3^{n+2} - 2n + 5}{(n+1)(n+2)}$$

$$\displaystyle \frac{3^{n+2} - 4n + 5}{(n+1)(n+2)}$$

$$\displaystyle \frac{3^{n+2} - 2n - 5}{(n+1)(n+2)}$$

None of these

Question 2

$$\dfrac{1}{4}$$

$$\dfrac{1}{7}$$

$$\dfrac{1}{3}$$

$$\dfrac{1}{5}$$

Solution

Question 3

$$1248$$

$$1370$$

$$1346$$

$$1452$$

$$1625$$

Solution

Question 4

$$\dfrac{33}{32}$$

$$-\dfrac{33}{32}$$

$$\dfrac{31}{32}$$

$$-\dfrac{33}{34}$$

Solution

Question 5

$$398$$

$$298$$

$$-398$$

$$198$$

Solution

Question 6

$$\dfrac{1}{2}$$

$$1$$

$$\dfrac{-1}{2}$$

$$-1$$

Solution

Question 7

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

$${ 2 }^{ 2n-2 }$$

$${ 2 }^{ 2n-3 }$$

$$\cfrac { { 2 }^{ 2n-2 } }{ \left( 2n-1 \right) ! } $$

Solution

Question 8

$$5580$$

$$5780$$

$$7789$$

$$1237$$

Solution

Question 9

$$2{ n }^{ 3 }+2{ n }^{ 2 }$$

$$\cfrac { 1 }{ 6 } \left( { n }^{ 3 }+3{ n }^{ 2 }+1 \right) $$

$$2{ n }^{ 3 }-2{ n }^{ 2 }$$

$$\cfrac { 1 }{ 6 } \left( { n }^{ 2 }-3{ n }+1 \right) $$

Solution

Question 10

$$(n+1)!$$

$$(n+2)!-1$$

$$n(n+1)!$$

none of these

Solution

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