Question

For each of these lists of integers, provide a simple formula orrule that generates the terms of an integer sequence that beginswith the given list. Assuming that your formula or rule is correct,determine the next three terms of the sequence.

b) 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, ...
c) 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011,...
d) 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, ...
f) 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425,...
g) 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, ...

Answer 1

b)
`T_(n)` = 3 + 4n

c)
`T_(n)` =
`n_(2)`

d)
`T_(n)` = n

f)
`T_(n)` =
`\pi ^(n)_(i=1) (2i - 1)`

g)
`T_(n)` =
`a^(n)` : =
`\left \{ {{a=0} if n is even \atop {a=1} if n is odd } \right.`

Explanation:

B) 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, ...

a = 7

difference = 4

this is an Arithmetic progression

`T_(n)` = a + (n - 1)d

= 7 + 4(n - 1)

`T_(n)` = 3 + 4n

the next three terms are 47, 51, 55

C) 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011,...

Given n >0 and n are integers

`T_(n)` =
`n_(2)`

the result is converted to base 2

the next three terms of the sequence are 1100, 1101, 1110

D) 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, ...

this is a square root of prime sequence where each number is repeated (2n -1) times

`T_(n)` = n

the next three terms of the sequence are
`7^(13), 9^(17), 13^(21)`.

F) 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425,...

`T_(n)` =
`\pi ^(n)_(i=1) (2i - 1)`

the next three term are

654729075, 1.374931058 x 10^10, 3.162341432 x 10^11

G) 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, ...

`T_(n)` =
`a^(n)` : =
`\left \{ {{a=0} if n is even \atop {a=1} if n is odd } \right.`

this is a sequence where each number is repeated (n) times

the next three terms are 0 in 6 places, 1 in 7places and 0 in 9places

Answer 2

b) next three terms of the sequence are 47, 51, 55

c) The next three terms are 1100, 1101, 1110

d) The next three items in the sequence are therefore, 8,8,8

f) The next three terms are 654729075, 13749310575, 316234143225

g) Therefore the next three items in the sequence are 0,0,0

Explanation:

b)7, 11, 15, 19, 23, 27, 31, 35, 39, 43, ...

the common difference, d =
`T_(2) - T_(1) = 11 - 7 = 4`

`a_(n) = a_(n-1) + 4`

43 + 4 = 47

47 + 4 = 51

51 + 4 = 55

next three terms of the sequence are 47, 51, 55

c) 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011,...

Each of these terms are binary numbers of 1,2,3,4........

The 1st next term is the 12th term = 1100₂

The second next term is the 13th term = 1101₂

The third next term is the 14th term = 1110₂

The next three terms are 1100, 1101, 1110

d) 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, ...

1 is repeated 1 time , 2 is repeated 3 times, 3 is repeated 5 times, 5 is repeated 7 times, the next number will sure be repeated 9 times( the repetition follows an odd number pattern)

To know the next number that will be repeated 9 times:

1+2 = 3

2+3 = 5

3 + 5 = 8

The pattern is that the previous summation adds to the preceding number

The next three items in the sequence are therefore, 8,8,8

f) 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425,...

The above sequence follows the pattern
`a_(n) = a_(n-1) (2n -1)\\`

The next three terms are the 10th, 11th and 12th terms

a₁₀ = 34459425((2*10)-1)

a₁₀ = 654729075

a₁₁ = 654729075((2*11)-1)

a₁₁ = 13749310575

a₁₂ = 13749310575((2*12)-1)

a₁₂ = 316234143225

The next three terms are 654729075, 13749310575, 316234143225

g) 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1,...

The sequence changes between 1 and 0. 6 zeros are expected next

Therefore the next three items in the sequence are 0,0,0