Mathematics savvy, ohoy!!!

[Svenskaflicka]

This is my homework for tomorrow:

200
SIGMA (2n+1)square
n=20

The question is: what is the total sum?

Now, me and my classmates tried with the formula for geometric number sequences:

Sn=a*k (raised in n) -1
. . . . . . k-1

And to get k we tried (2*21+1)square divided by (2*20+1)square, but that gave us 1,0999..., and when we tried to raise 1,1 in 200, our calculators went "ERROR:OVERFLOW" on us...

Am I missing something?

 

[elsol]

10816741

http://www.math.sc.edu/cgi-bin/sumcgi/calculator.pl

If I entered it correctly... according to my math memory and the calculator definition of sigma I entered.

sigma[20,200,(2*k+1)^2]

n = k because that's the what the calculator demanded I use.

 

[Jenny_Jackson]

You have to be careful how you write the expression here. You need parens to help the calculator because of the arithmetic priorities programmed into it.

http://i92.photobucket.com/albums/l27/Jenny_Jackson/ScreenShot128.jpg

 

[Guest]

<sigh> I miss MathGirl. She'd have given you an answer by now, a cunning insult and a gracious complement, while showing you a photo of herself and her hot aunt.

 

[Svenskaflicka]

<sigh> I miss MathGirl. She'd have given you an answer by now, a cunning insult and a gracious complement, while showing you a photo of herself and her hot aunt.

I miss her, too.

Thanks for the help, everyone. Tomorrow morning, when my head is clear, I'll try to see if I can figure out how you've reasoned with the math answer.

There might be something lost in translation between Swedish Mathematish and American Mathematish. It's hard to type the question properly without Greek Letters.


Tomorrow, then. I'm off to sleep now.

 

[Pure]

i get
10,816,741

agreeing with elsol


based on the formula for the sum of the first n squares

Sigma n**2 {running from 1 to n}= n(n+1)(2n+1)/6

what we are doing, if i read the problem correctly is trying to sum the ODD squares from 41 to 401.

So we need a formula for summing odd squares
{Sigma of} 1**2 + 3**2 + 5**2 ... (2p + 1)**2

Which formula is (unsimplified) (sum of the first p odd squares)
F#(p)= (2p+1)(2p+2)(4p+3)/6 - 4 (p)(p+1)(2p+1)/6

=(p+1)(2p+1)(2p+3)/3

The formula is to be evaluated for p=200 and for p=19

{F#(p) means F of p, NOT F times p.}

Then the answer is F#(200)-F#(19) = 10,827,401 - 10,660

=10,816,741
---

incidentally, summing squares is not summing a geometric progression, so the formula in the first posting is not applicable.

 

[Svenskaflicka]

You have to be careful how you write the expression here. You need parens to help the calculator because of the arithmetic priorities programmed into it.

http://i92.photobucket.com/albums/l27/Jenny_Jackson/ScreenShot128.jpg

??? When I wrote it exactly like you did, I got 10816741?

 

[Svenskaflicka]

Ah, no, now I get it. It's 2n PLUS 1 in square, not 2n MINUS 1.

Question is, will my teacher be happy with just the answer, or will he demand a calculation? And if so, will he accept a "well, that's what the claculator online said it would be..?"

 

[Pure]

i gave the formulas, not just the answer.

there is the sum of squares formula

from which you derive the sum of odd squares formula.(given)

this is evaluated for p=200, i.e., up to 401**2, giveing 10,815,060.

that is, 1**2 + 3**2... to +401**2.


however you want this sum to start at 41**2.

so you must substract off the sum of the odd squares up to 39**2

1**2 +3**2+ to+39**2

and that is 10,660.

hence the answer.

this only employed the simplest add, subtract, multiply and divide functions of a calculator, and indeed could be done with pencil and paper.

 

[Pure]

well, svenskaflicka, el,

did we get the right answer?