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a set of numbers is formed using a rule that the nth term is

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a set of numbers is formed using a rule that the nth term is [#permalink] New post 16 Feb 2005, 06:05
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a set of numbers is formed using a rule that the nth term is equal to 4n^2+1, for all positive n. in this set , the difference in value between the nth term and the next larger term is

a 4
b 12
c 3n^2
d 8n+4
e 8+n^2
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 [#permalink] New post 16 Feb 2005, 07:22
D it is


Tn= 4n^2 + 1

Tn+1 = 4(n+1)^2 +1=4n^2 +8n +5

==>>Tn+1- Tn= 4n^2 +8n+5-4n^2 +1 = 8n+4
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 [#permalink] New post 16 Feb 2005, 08:28
D.
i. (n)th term = 4n^2 + 1
ii. (n+1)th term=4(n+1)^2 +1
diff = {4(n^2+2n+1) +1}-(4n^2+1)
= 4n^2+8n+4+1-4n^2 - 1
= 8n+4
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 [#permalink] New post 16 Feb 2005, 08:44
D too...

The easiest way is to pick numbers from 1 to 3 and to check the answers:
when n = 1, result is 5
when n = 2, result is 17
when n = 3, result is 37

Difference is 12 between 1 and 2
Difference is 20 between 2 and 3

Just 8n+4 works.
  [#permalink] 16 Feb 2005, 08:44
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