How would I go about finding the sum of this series? Surely theres gotta be an easier way than throwing it all into the calculator. Thanks.

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- Oct 2nd 2007, 03:16 PM #1

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- Oct 2nd 2007, 03:37 PM #2
Can you follow this?

$\displaystyle \begin{array}{rcl}

\sum\limits_{k = 1}^{100} {\left( { - 1} \right)^{k + 1} k^2 } & = & \sum\limits_{k = 1}^{50} {\left[ {\left( {2k - 1} \right)^2 - \left( {2k} \right)^2 } \right]} \\

& = & \sum\limits_{k = 1}^{50} {\left[ {1 - 4k} \right]} \\

& = & 50 - 4\frac{{\left( {50} \right)\left( {51} \right)}}{2} \\

\end{array}

$

- Oct 2nd 2007, 03:47 PM #3

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- Oct 2nd 2007, 04:44 PM #4

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Hello, GeeDee!

Another approach . . .

We have: .$\displaystyle S \:=\:1^2-2^2+3^2-4^2+5^2-6^2 + \cdots + 99^2 - 100^2$

. . $\displaystyle = \;(1^2-2^2) + (3^2-4^2) + (5^2-6^2) + \cdots + (99^2-100^2) $

. . $\displaystyle = \;(1-2)(1+2) + (3-4)(3+4) + (5-6)(5+6) + \cdots + (99-100)(99+100) $

. . $\displaystyle = \;-1 - 2 - 3 - 4 - 5 - 6 - \cdots - 99 - 100$

This is an an arithmetic series with first term $\displaystyle a = \text{-}1$,

. . common difference $\displaystyle d = \text{-}1$, and $\displaystyle n = 100$ terms.

Its sum is: .$\displaystyle S \;=\;\frac{100}{2}\left[2(\text{-}1) + 99(\text{-}1)\right] \;=\;\boxed{-5050}$

- Oct 2nd 2007, 05:12 PM #5

- Oct 4th 2007, 01:45 PM #6

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Hello, Plato!

What kind of silly-ass question is that?

Are you taking lessons from Stapel and Mathman?

I've already left one math site because of this snotty attitude.

I taught math (successfully) for forty years. When students come to me for help,

I have never folded my arms and said, "If you can't show me your work, GET OUT!"

I have never believed in the Socratic approach . . . it takes far too long to DRAG

the reasoning and answers out of each and every student. I have found that

one clear, detailed solution can clear up the mystery for most of my students.

Certainly, many students simply want their homework done for them. Usually,

I can't tell from their initial posting. But if they reply with something immature like

"Are you SURE that's the right answer?", then I know I've been "had" and I won't

help that student again. Or I get no feedback from my solution and instead I get,

"Okay, how about this one...?", another loser ... lesson learned.

Do I want them learn? . . . Give me a break!

My life has been devoted to Teaching . . . not scolding or making snide remarks.

I give all student "the benefit of the doubt" at first. I do NOT treat them as

academic vagrants looking for a handout. I assume they are sincere in their quest,

whether it is a method, a hint, or a detailed explanation . . . which I assume

they will use for a template for the rest of their assignment.

Personally, I don't care how you teach or what you think. Just keep your little

snipes to yourself, okay?

- Oct 4th 2007, 04:35 PM #7
I do apologize to you for not knowing your philosophy of education. But I do have a few comments and observations.

Who are they? If they agree with me that theteaching method has never worked in mathematics, then I welcome their input.**show and tell**

I must say, I find that a shocking statement for any teacher.

I do respect that you think that, I infer that you a basically a trainer.

Is that a fair statement? Does that define the different in what community colleges do (i.e. train) and what universities do?

I find it very hard to think that giving a template is any more than training as opposed to learning.

Again, I do apologize for not realizing that we have such a fundamental disagreement on the difference in training and learning.

You and I clearly have a different view of mathematics education.

I pledge to you no more from me on your training efforts