calculus - sequences?

Question

i really suck at these problems. is there a trick or some kind of way to approach these things?

find the formula (a sub n) for the following sequences:

1. 2, 0, 2, 0, 2...

2. 1/2, 3/4, 7/8, 15/16, 31/32...

3. -1/4, 2/9, -3/16, 4/25, -5/36

sequences can be defined recursively: one or more terms are given explicitly; the remaining ones are then defined interms of their predecessors. give first 6 terms of sequence and then give nth term.

1. a1 = 1; a sub (n+1) = a sub n + 3n(n+1) + 1

2. a1 = 1; a sub (n+1) = 1/2(a sub n) + 1

i dont even know where to begin with these. please help. TIA

Answer 1

1. 1 + (-1)^n, n = 0, 1, 2, .... (-1)^n terms are useful for any sequence that has some alternation to it.

2. (2^n - 1) / 2^n, n = 1, 2, ... When you see 2,4,8,16, etc. you know a 2^n is involved.

3. (-1)^n * n / (n+1)^2, n = 1, 2, .... Learn to spot perfect squares like 1, 4, 9, 16, ...

Other tips: look how the number change from one to the next. Look at differences, or quotients. They can reveal a pattern that may not be apparent from the original sequence.

1. a1=1
a2 = 1 + 3*2*3 + 1 = 20
a3 = 20 + 3*3*4 + 1 = 57
a4 = 57 + 3*4*5 + 1 = 118
a5 = 118 + 3*5*6 + 1 = 209
a6 = 209 + 3*6*7 + 1 = 336
an = 1 + 3*sum i=1 to n i(i+1) + n - 7
= -6 + 3*sum i=1 to n i^2 + 3*sum i=1 to n i
= -6 + 3*n(n+1)(2n+1)/6 + 3n(n+1)/2 + n

How I got this one: notice a 1 is added every time, so that's where n comes in (because n 1's have been added in getting the nth term). The 3*sum i=1 to n i(i+1) comes from adding up the terms 3n(n+1) to get from an to an+1. The 1 comes from the initial value a1=1. The -7 comes from the fact that the 3n(n+1) + 1 part, used to obtain the other terms, is not included in a1, so I subtract that part off.

Also make use of the formulas

1+2+...+n = n(n+1)/2

1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6

2. a1=1
a2=1/2 + 1=3/2
a3=1/2*3/2 + 1 = 7/4
a4=1/2*7/4 + 1 = 15/8
a5=1/2*15/8 + 1 = 31/16
a6=1/2*31/16 + 1 = 63/32
an = (2^(n+1)-1)/2^n = 2 - 1/2^n.

Answer 2

Good tactics are either:
a) use trial-and-error to try candidate formulae and tweak your answer. Look for telltale factors of n, n^2, n^k, n!, 2^n, b^n. Often looking at the later terms first will reveal the pattern e.g. a factor of 36 is probably 6^2
or
b) look at the arithmetic difference or geometric ration between sucessive terms and spot a pattern.

1. 2, 0, 2, 0, 2...
a_n = 1 + (-1)^n

2. 1/2, 3/4, 7/8, 15/16, 31/32...
Note the telltale denominator is 2^n, and the numerator is 2^n -1
a_n = (2^n - 1)/(2^n) = 1 - 1/(2^n)

3. -1/4, 2/9, -3/16, 4/25, -5/36
Note the telltale denominator is (n+1)^2, and the numerator is n, with an alternating sign, i.e. also a (-1)^n factor:
a_n = ((-1)^n * n)/ (n+1)^2

> give first 6 terms of sequence and then give nth term.
This is easier, just plug in n=1 for the first term and crunch out the math, n=2 for second term etc...

1. a_1 = 1; a_(n+1) = a_n + 3n(n+1) + 1
Terms are a_1=1
a_2 = 1+1+ 3(2)(2+1) = 20
a_3 = 20+1 + 3(3)(3+1) = 57
a_4 = 57+1 + 3(4)(4+1) = 118
a_5 = 118+1 + 3(5)(5+1) = 209
a_6 = 209+1 + 3(6)(6+1) = 336

2. a_1 = 1; a_(n+1) = 1/2(a_n) + 1 ; (assuming the 2(a_n) is all on the denominator)
a_1 = 1
a_2 = 1/2(1) + 1 = 3/2
a_3 = 1/2(3/2) + 1 = 4/3
a_4 = 1/2(4/3) + 1 = 11/8
a_5 = 1/2(11/8) + 1 = 4/11 + 1 = 15/11
a_6 = 1/2(15/11) + 1 = 11/30 + 1 = 41/30

Answer 3

Take the difference or ratio between successive numbers and see if that has a pattern. The first one is pretty straightforward. The pattern is 2 then 0 repeating. For the next one, take the difference between the numbers.

3/4 - 1/2 = 1/4
7/8 - 3/4 = 1/8
15/16 - 7/8 = 1/16
31/32 - 15/15 = 1/32

Hopefully 1/4, 1/8, 1/16, 1/32 suggests a pattern to you.

That last one is more like what you would find on an IQ test. You have to look at the numerator and denominator seperately. The numerator series is 1,2,3,4,5 so that is pretty straight forward. The demoninator is 4,9,16,25,36 which is a list of squares, and the sign changes for each number.

Answer 4

9) The value of n indicates the term in the sequence, like n=2 is 1/3, n=3 is 1/5, which is shown in your problem. The general form for the sequence is: a(sub n) = 1/(2n-1) 19) I assume you are supposed to come up with some of the numbers of the sequence, as you simply stated the general form for some sequence. For that, simply plug n=1 into the general form to get the first value, n=2 to get the second value, and so on.

Answer 5

1. (1 - (-1)^n)
2. (2^n -1)/(2^n)
3. (-1)^n * n/(n+1)^2