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

a. 3, 6, 11, 18, 27, 38, 51, 66, 83, 102, . . .
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, . . .
e. 0, 2, 8, 26, 80,242,728,2186,6560,19682, . . .
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, . . .
h. 2, 4, 16, 256, 65536, 4294967296, . . .

A) Next 3 terms are; 123, 146, 171

B)Next 3 terms are; 47, 51, 55

C) Next 3 terms are; 1100, 1101, 1110

D)Next 3 terms are; 8, 8, 8

E) Next 3 terms are; 59048, 177146, 531440

F) Next 3 terms are; 654729075, 13749310575, 316234143225

G) Next 3 terms are; 0, 0, 0

H) Next 3 terms are; 18446744073709551616, 340282366920938463463374607431768211456, 115792089237316195423570985008687907853269984665640564039457584007913129639936

Step-by-step explanation:

A) 3, 6, 11, 18, 27, 38, 51, 66, 83, 102,...

In this sequence, if we inspect it starting from the first, we'll see that it increases by 3 from 3 to 6; increases by 5 from 6 to 11; increases by 7 from 11 to 18.

Thus we see a pattern where the next term after a previous one increases by 2n + 1.

So we can say;

An = A(n-1) + (2n - 1)

So for example, lets check the ones already done in the question ;

A3 = A2 + (2(3) - 1)

A3 = 6 + 5 = 11

Since it corresponds with the 3rd value in the sequence, let's now solve for the remaining 3 terms in the sequence;

So;

A11 = A10 + (2n-1)

A11 = 102 + (2(11) - 1) = 102 + 21 = 123

A12 = A11 + (2n-1)

A12 = 123 + (2(12) - 1) = 123 + 23 =146

A13 = A12 + (2n-1)

A13 = 146 + (2(13) - 1) = 146 + 25 = 171

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

We notice that each term increases by 4 from the previous term, thus like we derived from a above, we can see that;

An = An-1 + 4

A11 = 43 + 4 = 47

A12 = 47 + 4 = 51

A13 = 51 + 4 = 55

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

Inspecting the sequence, we can easily see that they are all in binary form and they increase in the exact form of how they are in binary.

Now, the next terms are the 12th 13th and 14th term.

When we convert any of the 1st to 11th term to binary, we get the eaxct same thing in the sequence.

So converting the 12th, 13th and 14th term to binary;

11th term;

12/2 = 6 R 0

6/2 = 3 R 0

3/2 = 1 R 1

2/1 = 0 R 1

So 12th term is 1100

Following same pattern, 13th will be 1101 and 14th will be 1110

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

Inspecting the sequence, we can see that ;

The numbers are repeated once, thrice, five times, seven times.

That means the next number will be repeated 9 times.

Also to determine the next number that's repeated 9 times, we can see that after 2,the next number is 2+1 = 3 and after 3 the next number is 3+2 = 5.

So we can conclude that after 5, the next number will be 5+3=8

Thus the next 3 numbers are 8,8,8

E)0,2,8,26,80,242,728,2186,6560,

19682

From the sequence, it's not as clear as previous ones.

Let's check 3 raised to the power of the position of the integer;

3¹, 3², 3³, 3⁴... To give 3,9,27,81

Inspecting the sequence, we can see that it follows the order of ;

An = [3^(n-1)] - 1

So let's check the second term in the sequence to confirm this pattern.

A2 = [3^(2-1)] - 1 = 3-1 = 2.

So the pattern is correct.

Thus ;

A11 = [3^(11-1)] - 1 = 3^(10) - 1 = 59048

A12 = [3^(12-1)] - 1 = 3^(11) - 1 = 177146

A13 = [3^(13-1)] - 1 = 3^(12) - 1 = 531440

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

Inspecting the above sequence, we can see that the 2nd term is the 1st term multiplied by 3; the 3rd term is the 2nd term multiplied by 5; the 4th term is the 3rd term multiplied by 7 and so on.

Thus, we can say;

An = (An-1) x (2n-1)

Thus;

A10 = A9 x (2(10) - 1) = 34459425 x 19 =654729075

A11 = A10 x (2(11)-1) = 654729075 x 21 = 13749310575

A12 = A11 x (2(12)-1) = 13749310575 x 23 = 316234143225

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

Looking at the sequence, we first see that it contains only numbers 1 and 0.

Secondly, we see that the number of times both are repeated increases.

Lastly we see that 0 is first repeated twice then 1 is repeated thrice, then 0 repeated four times and 1 repeated 5 times.

We can see that 1 and 0 alternate in terms of next to be repeated and that the next to be repeated is repeated one more time than the previous number repeated.

Thus; the next to be repeated is 0 and it will be repeated 6 times.

So the next 3 terms will be 0,0,0

H) 2, 4, 16, 256, 65536, 4294967296, .

Looking at the sequence, we can see that;

The next term is equal to the square of the previous term.

Thus, we can deduce that;

An = (An-1)²

A7 = (A6)² = 4294967296² = 18446744073709551616

A8 = (A7)² = 340282366920938463463374607431768211456

A9 = (A8)² = 115792089237316195423570985008687907853269984665640564039457584007913129639936

1.5%

step-by-step explanation:

rounding to the nearest thousandth would be 452,000