SOLUTION: HELP PLEASE!!! - Don't think I get it write the frist 6 terms of the sequence: 1.an = n^3-1 (first n is subscript) 2.an = (n-1)^2 (first n is subscrpt) Also- I need help w

Algebra ->  Sequences-and-series -> SOLUTION: HELP PLEASE!!! - Don't think I get it write the frist 6 terms of the sequence: 1.an = n^3-1 (first n is subscript) 2.an = (n-1)^2 (first n is subscrpt) Also- I need help w      Log On


   

Question 98709: HELP PLEASE!!! - Don't think I get it
write the frist 6 terms of the sequence:
1.an = n^3-1 (first n is subscript)
2.an = (n-1)^2 (first n is subscrpt)
Also- I need help with summation notation:
4+ 8+ +12+ 16 + 20 - write in summation notation
-7-8-9-10-11 in summation
5/6+6/7+7/8+8/9+... summation notation
Thanks so very much
Answer by Adam(64) About Me  (Show Source):
You can put this solution on YOUR website!
Sequence is just some (ordered) list of numbers(terms).The equation A%5Bn%5D+=+n%5E3-1 just tells you how to get n-th term. If you want to get first term, substitute 1 for n, so you get A%5B1%5D+=+1%5E3-1+=+1-1+=+0 , same with second term : substitute 2, in order to get : A%5B2%5D+=+2%5E3-1+=+8-1+=+7. Similiarly with A%5Bn%5D+=+%28n-1%29%5E2
1st : A%5B1%5D+=+%281-1%29%5E2+=+0%5E2+=+0
2nd : A%5B2%5D+=+%282-1%29%5E2+=+1%5E2+=+1
3rd : A%5B3%5D+=+%283-1%29%5E2+=+2%5E2+=+4
and so on...
Summation- you are just summing terms of sequence. For instance, if you have :
sum%28n%5E3-1%2Cn=1%2C6%29
, same like in the first problem, !but summed! So not 0 , 7 , 25 , 63 , but 0+7+25+63 = 95.(You are substituting through n=1 to n=6 and summing together.
or sum%28+n+%2C+n=0+%2C5%29 = 0+1+2+3+4+5 = 15
given 4+8+12+16+20 you can rewrite as sum%284%2An%2Cn+=+1+%2C+5%29
-7-8-9-10-10-11 as sum%28-n%2Cn=7%2C+11%29
and finally 5/6+6/7+7/8+8/9+... as sum%28n%2F%28n%2B1%29%2Cn+=+5+%2C+infinity%29

-method of rewriting into summation notation is always based on some preliminary thought, as multiples of 4 etc.. so go through it and give some thought about it...