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 December 15th, 2006, 01:34 PM #1 Newbie   Joined: Dec 2006 Posts: 28 Thanks: 0 Term of the geometric Find the sum of the first 10 terms of the geometric series whose first term is -6 and whose common ratio is -2 December 15th, 2006, 04:57 PM #2 Guest   Joined: Posts: n/a Thanks: Why can't you just use the formula for a geometric series?: a/(1-r) (-6)/(1-(-2))=-2
December 16th, 2006, 03:27 AM   #3
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Quote:
 Why can't you just use the formula for a geometric series?: a/(1-r) (-6)/(1-(-2))=-2
Yes, but that's the sum of an infinite quantity of terms, and in this case, the sum doesn't equal
-2 if r=-2, because if r>=1, a geometric series diverges to infinity, and the a/(1-r) formula doesn't work. You have to prove that the series converges first before you can find out what it converges to. In this case, the series would diverge by the nth term divergence test, since the lim n->00 of (-6)(-2)^n =00 . December 16th, 2006, 07:44 AM #4 Newbie   Joined: Dec 2006 Posts: 28 Thanks: 0 Thank guys very much.. just asking from with lesson is that. i mean what kind of problem is this January 29th, 2007, 02:13 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,934 Thanks: 2208 (-6 + 12) + (-24 + 48) + (-96 + 192) + (-384 + 768) + (-1536 + 3072) = 6 + 24 + 96 + 384 + 1536 = 2046 Sum of n terms = a(1 - r^n)/(1-r) = -6(1 - (-2)^n)/(1 - (-2)) (for the series in the above problem) = -2(1 - (-2)^n) = -(-2)^(n+1) - 2 = 2^(n+1) - 2 if n is even; - 2^(n+1) - 2 if n is odd. Tags geometric, term Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post bongantedd Algebra 2 February 26th, 2014 04:38 PM kingkos Algebra 2 December 21st, 2012 07:19 PM rnck Applied Math 3 August 23rd, 2012 07:17 PM forcesofodin Real Analysis 4 August 14th, 2010 05:50 PM rnck Real Analysis 3 December 31st, 1969 04:00 PM

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