the nth term

[logistic_guy]

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.

\(\displaystyle \bold{a)} \ 1,0,1,1,0,0,1,1,1,0,0,0,1,\cdots\)
\(\displaystyle \bold{b)} \ 1,2,2,3,4,4,5,6,6,7,8,8,\cdots\)
\(\displaystyle \bold{c)} \ 1,0,2,0,4,0,8,0,16,0,\cdots\)
\(\displaystyle \bold{d)} \ 3,6,12,24,48,96,192,\cdots\)
\(\displaystyle \bold{e)} \ 15,8,1,-6,-13,-20,-27,\cdots\)
\(\displaystyle \bold{f)} \ 3,5,8,12,17,23,30,38,47,\cdots\)
\(\displaystyle \bold{g)} \ 2,16,54,128,250,432,686,\cdots\)
\(\displaystyle \bold{h)}\ 2,3,7,25,121,721,5041,40321,\cdots\)

 

[logistic_guy]

\(\displaystyle \bold{a)}\)

The pattern is:
\(\displaystyle 1,0\)
\(\displaystyle 1,1,0,0\)
\(\displaystyle 1,1,1,0,0,0\)
\(\displaystyle 1,\)1\(\displaystyle ,\)1\(\displaystyle ,\)1\(\displaystyle ,0,0,0,0\)

 

[logistic_guy]

\(\displaystyle \bold{b)}\)

The pattern is:
\(\displaystyle 1,2,2\)
\(\displaystyle 3,4,4\)
\(\displaystyle 5,6,6\)
\(\displaystyle 7,8,8\)
9\(\displaystyle ,\)10\(\displaystyle ,\)10

 

[logistic_guy]

\(\displaystyle \bold{c)}\)

The pattern is:
\(\displaystyle 2^0,0\)
\(\displaystyle 2^1,0\)
\(\displaystyle 2^2,0\)
\(\displaystyle 2^3,0\)
\(\displaystyle 2^4,0\)
2\(\displaystyle ^5, \ \)0
2\(\displaystyle ^6\)\(\displaystyle ,0\)

Or

\(\displaystyle 1,0\)
\(\displaystyle 2,0\)
\(\displaystyle 4,0\)
\(\displaystyle 8,0\)
\(\displaystyle 16,0\)
32\(\displaystyle , \ \)0
64\(\displaystyle ,0\)

 

[logistic_guy]

\(\displaystyle \bold{d)}\)

A simple formula is:
\(\displaystyle 3 \times 2^{n-1}\)

The \(\displaystyle 8^{\text{th}}, 9^{\text{th}}\) and \(\displaystyle 10^{\text{th}}\) terms are:

\(\displaystyle 3 \times 2^{8-1}\)
\(\displaystyle 3 \times 2^{9-1}\)
\(\displaystyle 3 \times 2^{10-1}\)

Or

384
768
1536

 

[logistic_guy]

\(\displaystyle \bold{e)}\)

I created this little nice formula.
\(\displaystyle 22 - 7n\)

\(\displaystyle 22 - 7(1) = 15\)
\(\displaystyle 22 - 7(2) = 8\)
\(\displaystyle 22 - 7(3) = 1\)
\(\displaystyle 22 - 7(4) = -6\)
\(\displaystyle 22 - 7(5) = -13\)
\(\displaystyle 22 - 7(6) = -20\)
\(\displaystyle 22 - 7(7) = -27\)
\(\displaystyle 22 - 7(8) = \) -34
\(\displaystyle 22 - 7(9) = \) -41
\(\displaystyle 22 - 7(10) = \) -48

 

[logistic_guy]

\(\displaystyle \bold{f)} \ 3,5,8,12,17,23,30,38,47,\cdots\)

I created this little dirty formula.
\(\displaystyle \frac{n^2}{2} + \frac{n}{2} + 2\)

\(\displaystyle \frac{(1)^2}{2} + \frac{(1)}{2} + 2 = 3\)

\(\displaystyle \frac{(2)^2}{2} + \frac{(2)}{2} + 2 = 5\)

\(\displaystyle \frac{(3)^2}{2} + \frac{(3)}{2} + 2 = 8\)

\(\displaystyle \frac{(4)^2}{2} + \frac{(4)}{2} + 2 = 12\)

\(\displaystyle \frac{(5)^2}{2} + \frac{(5)}{2} + 2 = 17\)

\(\displaystyle \frac{(6)^2}{2} + \frac{(6)}{2} + 2 = 23\)

\(\displaystyle \frac{(7)^2}{2} + \frac{(7)}{2} + 2 = 30\)

\(\displaystyle \frac{(8)^2}{2} + \frac{(8)}{2} + 2 = 38\)

\(\displaystyle \frac{(9)^2}{2} + \frac{(9)}{2} + 2 = 47\)

\(\displaystyle \frac{(10)^2}{2} + \frac{(10)}{2} + 2 = \) 57

\(\displaystyle \frac{(11)^2}{2} + \frac{(11)}{2} + 2 = \) 68

\(\displaystyle \frac{(12)^2}{2} + \frac{(12)}{2} + 2 = \) 80