# help with arithmetic and geometric sequences

• Dec 6th 2010, 03:36 AM
maxi
help with arithmetic and geometric sequences
hi folks.

i'm having a bit of a mental block here. i'm studying series for my exam next week and there's something i cant quite get. in an arithmetic progression each subsequent term has a common difference and a geometric progression has a common ratio, i'm absolutely fine with this. this is my problem, take the squares for example;

1, 4, 9, 16, 25, ... and so on.

i cant see what the common ratio is here. another example;

(1 × 2) + (2 × 5) + (3 × 8) + · · · to n.

i'm trying to write that one in sigma notation but i just cant see what the common ratio is, according to my notes it should be k(3k-1). is there process i can use to figure this out or am i just supposed to "see" the pattern? any help here would be great because when it comes to finding the sum of a geometric series i'm completely done in so thanks in advance.
• Dec 6th 2010, 03:50 AM
yeKciM
Quote:

Originally Posted by maxi
hi folks.

i'm having a bit of a mental block here. i'm studying series for my exam next week and there's something i cant quite get. in an arithmetic progression each subsequent term has a common difference and a geometric progression has a common ratio, i'm absolutely fine with this. this is my problem, take the squares for example;

1, 4, 9, 16, 25, ... and so on.

i cant see what the common ratio is here. another example;

(1 × 2) + (2 × 5) + (3 × 8) + · · · to n.

i'm trying to write that one in sigma notation but i just cant see what the common ratio is, according to my notes it should be k(3k-1). is there process i can use to figure this out or am i just supposed to "see" the pattern? any help here would be great because when it comes to finding the sum of a geometric series i'm completely done in so thanks in advance.

it's more to see patterns :D that come with practice :D

but it's just logic :D for that second of yours ...

$1\cdot2+2\cdot5+3\cdot8+....$

as you see first number is changing by the one :D so it will be n* od k* ... now let's look at the second ... how can u write 2+5+8+11 ? as you see it's next one always bigger from the one before by 3 ... so you know that first is n or k or whatever ... so second is 3k and you now do -1 because of the first first member :D to get that what you need :D
because without -1 youl get 1*3 + 2*6 +3*9 and so on ... :D:D:D so it must be k(3k-1)

now go back at first example :D

$1+4+9+16+...$

so it's .... for just k.... will be 1+2+3+4+... do you see pattern similar to yours ? $k^2$ perhaps ? :D

k*k = 1*1 +2*2+3*3+4*4+.... :D

$k^2 = 1+ 4+ 9+ 16 +25 + ....$

you can try like this ... you know that your series change by one :D ? ok :D so whatever you have you can write as 1* something + 2* something(the same) + ... and then compare with that what you need and then just solve what is that something that when multiplied with number of the member will give you your sequence ... sorry for bad explaining ... few problems with English :D:D:D:D
• Dec 6th 2010, 07:59 AM
maxi
ok thank you. so for progressions such as these, to find the common ratio, you're saying it is necessary to introduce a variable (n or k etc)? i thought the common ratio had to be a known constant hence my confusion, eg;

3, 9, 27, 81, ...

this progression is simple. the common ratio = 3 and there is no need to use a variable. is my thinking here correct?
• Dec 6th 2010, 10:02 AM
yeKciM
Quote:

Originally Posted by maxi
ok thank you. so for progressions such as these, to find the common ratio, you're saying it is necessary to introduce a variable (n or k etc)? i thought the common ratio had to be a known constant hence my confusion, eg;

3, 9, 27, 81, ...

this progression is simple. the common ratio = 3 and there is no need to use a variable. is my thinking here correct?

perhaps i'm a bit lost in translating ... but ...

$1+4+9+16+25 + .... = \sum _{n=1} ^{\infty} n^2$

that's why there is "new" variable... :D to write it in series .... because sum changes by n or whatever .... meaning members :D