Show that, when terms of the series are grouped in pairs, an arithmetic series is formed.
Follow Math Help Forum on Facebook and Google+
Originally Posted by SengNee Show that, when terms of the series are grouped in pairs, an arithmetic series is formed. Grouping the terms gives pairs of the form: so: RonL
Last edited by CaptainBlack; January 21st 2008 at 12:20 AM. Reason: Correct (2k)^2 typo
Originally Posted by SengNee Show that, when terms of the series are grouped in pairs, an arithmetic series is formed. Originally Posted by CaptainBlack Grouping the terms gives pairs of the form: so: RonL So to explicitly address the show bit, now that the hard work's been done: therefore ....
Originally Posted by CaptainBlack Grouping the terms gives pairs of the form: so: RonL I don't understand this part...
Originally Posted by SengNee I don't understand this part... It's a shorthand way of writing . If you haven't met this notation before, don't worry about it. If you want to research it a bit, it's called summation or sigma notation.
Hello, SengNee! Here's a rather primitive approach . . . Show that, when terms of the series are grouped in pairs, an arithmetic series is formed. Group in pairs: . . . . . . . . . . This is an arithmetic series with first term and common difference
Originally Posted by mr fantastic It's a shorthand way of writing . If you haven't met this notation before, don't worry about it. If you want to research it a bit, it's called summation or sigma notation. I think the first one is false, the second one is correct. Because for the first one, when , For the second one, when , Isn't it?
Last edited by SengNee; January 20th 2008 at 11:18 PM.
Originally Posted by Soroban Hello, SengNee! Here's a rather primitive approach . . . Group in pairs: . . . . . . . . . . This is an arithmetic series with first term and common difference I do like this. Is it correct also?
View Tag Cloud