SOLUTION: find a formula for t_n for the following: 0, -5, -10, -15... 1, -3, 9, -27... 2, 5, 10 ,17...

find a formula for tn for the following: 

0, -5, -10, -15... 

Test to see whether it is an arithmetic or geometric sequence
or neither:

If it is arithmetic, then t2-t1 = t3-t2 = t4-t3 = d
If it is geometric, then t2/t1 = t3/t2 = t4/t3 = r

It is not geometric since we can't divide -5 by 0.

Test for arithmetic sequence:

t2-t1 = (-5)-(0) = -5
t3-t2 = (-10)-(-5) = -10+5 = -5
t4-t3 = (-15)-(-1) = -15+10 = -5

These are all equal so it is an arithmetic sequence with
common difference d = -5

The formula for tn is

tn = t1 + (n-1)d

tn = 0 + (n-1)(-5)

tn = -5(n-1)

tn = -5n + 5

tn = 5 - 5n 

--------------------------------

1, -3, 9, -27... 

Test to see whether it is an arithmetic or geometric sequence
or neither:

If it is arithmetic, then t2-t1 = t3-t2 = t4-t3 = d
If it is geometric, then t2/t1 = t3/t2 = t4/t3 = r

Test for arithmetic sequence:

t2-t1 = (-3)-(1) = -4
t3-t2 = (9)-(-3) = 9+3 = 12
t4-t3 = (-27)-(9) = -36

These are not equal so it is not an arithmetic sequence.

Test for geometric sequence:

t2/t1 = (-3)/(1) = -3
t3/t2 = (9)/(-3) = -3
t4/t3 = (-27)/(9) = -3

These are all equal so it is a geometric sequence with
common ration r = -3

The formula for tn is

tn = t1rn-1

tn = 1(-3)n-1

tn = (-3)n-1

------------------------------

2, 5, 10 ,17...

Test for arithmetic sequence:

t2-t1 = (5)-(2) = 3
t3-t2 = (10)-(5) = 5
t4-t3 = (17)-(10) = 7

These are not equal so it is not an arithmetic sequence.

Test for geometric sequence:

t2/t1 = (5)/(2) = 3/2
t3/t2 = (10)/(5) = 2/3
t4/t3 = (17)/(10) = 17/10

These are not equal so it is not an geometric sequence either.

When it is neither, we have to use our own intuition to see
if we can relate it to some well-known sequence.

If you look at the numbers carefully,

2, 5, 10 ,17...

and think about the common sequence of squares of integers:

1, 4, 9, 16,...

which has formula tn = n2

you see that every number in 2, 5, 10 ,17... is just 1 more
than the corresponding term in the sequence of squares of
integers, so since the formula for tne sequence of squares 
is tn = n2, then the formula for the sequence every term of
which is 1 more than the corresponding term of the sequence of
squares of integers is

tn = n2 + 1

Edwin