Sequences and Series Test

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Sequences and Series Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

Let $S_1=\displaystyle \sum_{K=1}^{n}K,S_2=\sum_{K=1}^{n}K^2$ and $S_3=\displaystyle \sum_{K=1}^{n}K^3$ , then $\dfrac{S_{1}^{4}S_{2}^{2}-S_{2}^{2}S_{3}^{2}}{S_{1}^{2}+S_{3}^{2}}$ is equal to

A
$4$

B
$2$

C
$1$

D
$0$

Solution

$\displaystyle S_{1}= \sum k = \frac{n(n+1)}{2}, S_{2}=\sum k^{2}=\frac{n(n+1)(2n+1)}{6}, S_{3}=\sum k^{3} = \frac{n^{2}(n+1)^{2}}{4}$

$\displaystyle \frac{S_{1}^{4}S_{2}^{2}-S_{2}^{2}S_{3}^{2}}{S_{1}^{2}+S_{3}^{2}}$ $\displaystyle = \frac{ \dfrac{n^{4}(n+1)^{4}}{2^{4}}\,.\dfrac{n^{2}(n+1)^{2}(2n+1)^{2}}{36}-\dfrac{n^{2}(n+1)^{2}(2n+1)^{2}}{36}\,.\dfrac{n^{4}(n+1)^{4}}{4^{2}} }{S_{1}^{2}+S_{3}^{2}}$

$\displaystyle= 0$

$\therefore$ option D is correct
Question 2
1 / -0

The first three terms of a sequence are 3,3,6 and each term after the second is the sum of two terms preceding it, the 8th term of the sequence is

A
15

B
24

C
39

D
63

Solution

The given series is 3 times the Fibonacci series so if the 8th of Fibonacci series is 21 then 8th term of this series would be 63
Question 3
1 / -0

Choose the most appropriate option which follows the pattern

A

B

C

D
None of the above
Question 4
1 / -0

What is the next term to this series, 2, 3, 7, 16, 32, and 57...?

A
94

B
93

C
92

D
95

Solution

Let's analyse the series to get the answer,

2 + $(1)^2$ = 3
3 + $(2)^2$ = 7
7+ $(3)^2$ = 16
16 + $(4)^2$ = 32
32 + $(5)^2$ = 57

So, Next term $\rightarrow 57 + (6)^2 = 93$

Thus, B is the correct answer.
Question 5
1 / -0

Find the next term in the sequence.
$117, 389, 525, 593, 627 ?$

A
644

B
640

C
634

D
630

E
none of these

Solution
Question 6
1 / -0

The value of the expression
$\left(1 + \dfrac{1}{\omega}\right) \left(1 + \dfrac{1}{\omega^2}\right) + \left(2 + \dfrac{1}{\omega}\right) \left(2 + \dfrac{1}{\omega^2}\right) + \left(3 + \dfrac{1}{\omega^2}\right) \left(3 + \dfrac{1}{\omega^2}\right) +.............+ \left(n + \dfrac{1}{\omega}\right) \left(n + \dfrac{1}{\omega^2}\right)$

( $\omega$ is the root of unity) is

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

B
$\dfrac{n(n^2 - 2)}{3}$

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

D
None of these

Solution

$a_n = \left ( n +\dfrac {1}{\omega } \right )\left ( n +\dfrac {1}{\omega^2} \right )$

$\sum a_n= \sum \left ( n+\dfrac {1}{\omega } \right )\left ( n+\dfrac {1}{\omega^2 } \right )$

$= \sum \left ( \dfrac {\omega n +1}{\omega } \right )\left ( \dfrac {\omega^2 n +1}{\omega^2 } \right )$

$= \sum \dfrac {1}{\omega^3}(\omega^3 n^2 +\omega n + \omega^2 n+1)$

$= \sum n^2 + (\omega+ \omega^2)n+1 \,\,\,\, .............. [\omega^3=1]$

$= \sum n^2 +(-1)n+1 \,\,\,\, ................ [1+\omega+\omega^2=0]$

$= \sum n^2 - n +1$

$= \sum n^2 - \sum n + \sum 1$

$= n\left [ \dfrac {(n+1)(2n+1)}{6} -\dfrac {(n+1)}{2} +1\right ]$

$= n \left [ \dfrac {2n^2+2n+n+1-3n -3+6}{6} \right ]$

$= n \left [ \dfrac {2n^2+4}{6} \right ]$

$= \dfrac {n(n^2+2)}{3}$
Question 7
1 / -0

Find the number suitable for blank in given series.
1,2,3,5,8,13,21,___,55

A
35

B
33

C
36

D
34

Solution

The required number can be found by adding two numbers before it .
Examples : 3 can be found by adding two number before it.
i.e => (1+2) = 3
Similarly 5  can be found by adding two number before it.
i.e => (2+3) = 5 and so on.
In the same way the missing number can be found by adding  two number before it .
i.e => 13 + 21 = 34
The required number is 34. (Ans)
Question 8
1 / -0

If D=4 and COVER = 63, then BASIC is equal to

A
49

B
50

C
34

D
55

Solution

In the alphabets D is 4th.
So, A = 1, B = 2 and C =3
So, COVER = 3+15+22+5+18 = 63
Therefore BASIC = 2+1+19+9+3 = 34
Question 9
1 / -0

If $5\times 9=144; 7\times 8=151:4\times 6=102,$ then $2\times 5=?$

A
$73$

B
$77$

C
$37$

D
$97$

Solution

$5\times 9=\underset{(5+9)}{14}\underset{(9-5)}{4}$ ; $7\times 8=\underset{(7+8)}{15}\underset{(8-7)}{1}$ So, $2\times 5=\underset{(2+5)}{7}\underset{(5-3)}{3}$
Question 10
1 / -0

Observe the given pattern
$4\times 0+1=01$
$4\times1+2=06$
$4\times 2+3=11$
$4\times 3+4=16$
Then the value of $4\times8+9$ is -----

A
$43$

B
$42$

C
$41$

D
$40$

Solution

If we carefully observe the pattern
in first row it shows four multiply with zero addition one equals one
same for other pattern too
so answer of $4\times 8+9 = 41$

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Re-Attempt Weekly Quiz Competition

Sequences and Series Test - 16

Result

Sequences and Series Test - 16

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SHARING IS CARING

Question 1

444

222

111

000

Solution

Question 2

15

24

39

63

Solution

Question 3

None of the above

Question 4

94

93

92

95

Solution

Question 5

644

640

634

630

none of these

Solution

Question 6

n(n2+2)3\dfrac{n(n^2 + 2)}{3}3n(n2+2)​

n(n2−2)3\dfrac{n(n^2 - 2)}{3}3n(n2−2)​

n(n2+1)3\dfrac{n(n^2 + 1)}{3}3n(n2+1)​

None of these

Solution

Question 7

35

33

36

34

Solution

Question 8

49

50

34

55

Solution

Question 9

737373

777777

373737

979797

Solution

Question 10

434343

424242

414141

404040

Solution

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Question Analysis

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$4$

$2$

$1$

$0$

$\dfrac{n(n^2 + 2)}{3}$

$\dfrac{n(n^2 - 2)}{3}$

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

$73$

$77$

$37$

$97$

$43$

$42$

$41$

$40$