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 February 27th, 2012, 04:01 PM #1 Newbie   Joined: Jan 2012 Posts: 28 Thanks: 0 Find the sum of the first 50 terms Find the sum of the first 50 terms of the following sequence: 1 + 3 - 5 + 7 + 9 - 11 + 13 + 15 - 17 + ...
 February 27th, 2012, 05:36 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Find the sum of the first 50 terms We have the sum S given by: $S=1+3-5+7+9-11+13+\cdots-89+91+93-95+97+99$ $S=1+3+(-5+7)+9+(-11+13)+\cdots+(-89+91)+93+(-95+97)+99$ $S=1+(3+2+9+2+\cdots+2+93+2+99+2)-2$ $S=-1+\sum_{k=0}^{16}$$6k+5$$=901-1=900$
 February 27th, 2012, 05:45 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Find the sum of the first 50 terms We could also write: S = 1 + 3 - 5 + 7 + 9 - 11 + ... - 89 + 91 + 93 - 95 + 97 + 99 S = 99 + 97 - 95 + 93 + 91 - 89 + ... - 11 + 9 + 7 - 5 + 3 + 1 Adding we get: 2S = 16(100 + 100 - 100) + 100 + 100 = 1800 S = 900
 February 27th, 2012, 05:56 PM #4 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 2 Re: Find the sum of the first 50 terms Mark looks like Carl Friedrich Gauss to me .
February 27th, 2012, 06:09 PM   #5
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Re: Find the sum of the first 50 terms

[attachment=0:21x2xpqg]gaussplate.jpg[/attachment:21x2xpqg]
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February 27th, 2012, 07:53 PM   #6
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Re: Find the sum of the first 50 terms

Quote:
 Originally Posted by MarkFL We could also write: S = 1 + 3 - 5 + 7 + 9 - 11 + ... - 89 + 91 + 93 - 95 + 97 + 99 S = 99 + 97 - 95 + 93 + 91 - 89 + ... - 11 + 9 + 7 - 5 + 3 + 1
Similarly; looking at it as 2 series:

1 + 3 + 5 + 7 + ....... + 95 + 97 + 99 = 2500

2(-5 - 11 - 17 ....... - 89 - 95) = -1600

Throwing that mess in the arithmetic series formula:
50(1 + 99) / 2 - 2[16(5 + 95) / 2] = 900

February 28th, 2012, 03:38 AM   #7
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Re: Find the sum of the first 50 terms

Hello, stevecowall!

And yet another approach . . .

Quote:
 Find the sum of the first 50 terms of the following sequence: [color=beige]. . [/color]1 + 3 - 5 + 7 + 9 - 11 + 13 + 15 - 17 + ...

$S \;=\;(1\,+\,3\,-\,5)\:+\:(7\,+\,9\,-\,11)\:+\:(13\,+\,15\,-\,17)\:+\:\cdots\:+\:(91\,+\,93\,-\,95) \:+\:97\:+\:99$

$S \;=\;\underbrace{-1\,+\,5\,+\,11\,+\,\cdots\,+\,89}_{\text{arithmeti c series}}\:+\:97\:+\:99$

$\text{The series has first term }a = -1,\text{ common difference }d = 6,\text{ and }n = 16\text{ terms.}
\;\;\;\;\;\text{Its sum is: }\:\frac{16}{2}\big[2(-1)\,+\,15(6)\big] \:=\:704$

$\text{Therefore: }\:S \;=\;704\,+\,97\,+\,99 \;=\;900$

 February 28th, 2012, 08:40 AM #8 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 The mean of the first 2 terms is 2. The mean of the first 5 terms is 3. The mean of the first 8 terms is 4. ... The mean of the first 50 terms is 18, so their sum is 900.
February 28th, 2012, 12:59 PM   #9
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Re:

Quote:
 Originally Posted by skipjack The mean of the first 2 terms is 2. The mean of the first 5 terms is 3. The mean of the first 8 terms is 4. ... The mean of the first 50 terms is 18, so their sum is 900.
OH WOW! This is nice! Do you have 'skipjack' on your license plate?

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# algebra find the sum of the 50 terms of the arithemetic sequese 8,10,12,14,16

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