Stuck on seemingly easy sequence problem...

[eric_f]

The problem states:

Calculate the first six terms of the following sequence (start with n=1):

c(sub n) = n+(n+1)+(n+2)+...+(2n)

I am stuck at the beginning! How do I know where to fill in terms such as (n+3) and where to stop filling in terms and letting (2n) be the last? I'm probably overthinking this...

I started and ended up with this:

1+(1+1)+(2(1))
2+(2+1)+(2(2))
3+(3+1)+(3+2)+(2(3))
4+(4+1)+(4+2)+(4+3)+(2(4))
5+(5+1)+(5+2)+(5+3)+(5+4)+(2(5))
6+(6+1)+(6+2)+(6+3)+(6+4)+(6+5)+(2(6))

How wrong is that?

 

[stapel]

Calculate the first six terms of the following sequence (start with n=1):

c(sub n) = n+(n+1)+(n+2)+...+(2n)

I am stuck at the beginning! How do I know where to fill in terms such as (n+3) and where to stop filling in terms and letting (2n) be the last?

I'm guessing that you've never before encountered an ellipsis (the "dot-dot-dot" in the middle), so you're not familiar with what it means. You can read a general description here. In this particular context, it means "and the pattern continues in the same way" for a while, until it reaches a part that's interesting, or somehow denotes the end.

In this particular example, the rule appears to be "start with 'n', then count off by ones, and don't stop until you reach the value '2n'; then add all those values".

In your "answers, you just did the first two or three terms, and then the last, without including any of the rest of the terms (or including extras, if the list should have had fewer terms) so, naturally, the results didn't quite pan out. Now, try using the rule they gave you, and see where that leads.

 

[eric_f]

Thanks, what about these:

1+(1+1)
2+(2+1)+(2+2)
3+(3+1)+(3+2)+(3+3)
4+(4+1)+(4+2)+(4+3)+(4+4)
5+(5+1)+(5+2)+(5+3)+(5+4)+(5+5)
6+(6+1)+(6+2)+(6+3)+(6+4)+(6+5)+(6+6)

That starts with n, and adds one until the value of 2n is reached. Am I on the right track? I know I have to do the arithmetic, but that's the easy part.

 

[stapel]

Have they given you any formulas? Like for the sum of the first "n" whole numbers (so 1 + 2 + ... + n)?

 

[eric_f]

Sorry for the delay, here's the question verbatim:

(1) Compute the first six terms of the following sequences (start with n = 1):

(c) c(sub n) = n + (n + 1) + (n + 2) + . . . + (2n) (think carefully about this one!)

I end up with:

c1=1+(1+1)=3
c2=2+(2+1)+(2+2)=9
c3=3+(3+1)+(3+2)+(3+3)=18
c4=4+(4+1)+(4+2)+(4+3)+(4+4)=30
c5=5+(5+1)+(5+2)+(5+3)+(5+4)+(5+5)=45
c6=6+(6+1)+(6+2)+(6+3)+(6+4)+(6+5)+(6+6)=63

 

[stapel]

Okay; so why are you thinking these are wrong? What does the solution manual (or answer in the back of the book) say? Thank you!

 

[pka]

Sorry for the delay, here's the question verbatim:

(1) Compute the first six terms of the following sequences (start with n = 1):

(c) c(sub n) = n + (n + 1) + (n + 2) + . . . + (2n) (think carefully about this one!)

I end up with:

c1=1+(1+1)=3
c2=2+(2+1)+(2+2)=9
c3=3+(3+1)+(3+2)+(3+3)=18
c4=4+(4+1)+(4+2)+(4+3)+(4+4)=30
c5=5+(5+1)+(5+2)+(5+3)+(5+4)+(5+5)=45
c6=6+(6+1)+(6+2)+(6+3)+(6+4)+(6+5)+(6+6)=63

\(\displaystyle \displaystyle{c_n} = \sum\limits_{k = 0}^n {\left( {n + k} \right)} = \sum\limits_{k = 0}^n {\left( n \right)} + \sum\limits_{k = 1}^n {\left( k \right)} = \left( {n + 1} \right)n + \dfrac{{n(n + 1)}}{2} = \dfrac{{3n(n + 1)}}{2}\)