Summand part in summation notation

[aisha]

I need to write the following series in summation notation

1) 1+3+5+7+9+11 SUMMAND (2k-1)? is this right?

2) 4+6+8+10+12+12+16+18 (2k+2)? is this right?

Have I got it?

 

[dextercioby]

Okay.What does these symbols mean
\sum_{k=1}^{n} k ?

Daniel.

 

[aisha]

\sum_{k=1}^{n} k

ok the n= the last number in the series

k=1 the one is the first number in the series

k is the summand its used to get the terms in the series u input number k through n to get the series

 

[aisha]

ok for the first series i posted i got the summand to be (2k-1) with a 6 over the sigma and for the second series I got (2k+2) as the summand with 18 over the sigma, is this correct?

 

[dextercioby]

Perfect,then u agree it means just:1+2+...+k+...+n ...?

Okay.Now imagine how would your first sum would look like...You already did...Great.

\sum_{k=0}^{5} (2k+1)

Agree...?

Daniel.

P.S.For some reason,we prefer the "+" for the general form of an odd #.

 

[dextercioby]

Nope,not really.U see,the last term must coincide with the value of the general term when evaluated with the superior value:
\sum_{k=0}^{5} (2k+1)=...+11

11=(2k+1)|_{k=5}...

Okay...?

Daniel.

 

[aisha]

\sum_{k=0}^{5} (2k+1)

ok so this is the only answer for the first series?

\sum_{k=4}^{18} (2k+2) and is this answer correct for the second series?

 

[dextercioby]

Okay,true.Have your way,it's basically the same thing...

Daniel.

 

[dextercioby]

\sum_{k=0}^{5} (2k+1)

ok so this is the only answer for the first series?

\sum_{k=4}^{18} (2k+2) and is this answer correct for the second series?

No,no,as i just said,your answer is true as well.Just for the first.For the second,the "k" should go from 1------>8.

Daniel.

 

[aisha]

show me how the second one looks I don't understand from 2-8?

 

[dextercioby]

\sum_{k=1}^{8}(2k+2) produces the same terms as the ones you had.

Daniel.

 

[aisha]

how come 2 and 8 are right the series didnt start with 2 or end at 8

 

[dextercioby]

It's "+1" --------->"+8".It was a tiny mistake.I've edited my posts.

(2k+2)|_{k=1}=2\times 1+2=4
-----------------
(2k+2)|_{k=8}=2\times 8+2=18

Okay?

Daniel.