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 $$4\times 8+9 = 41$$