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Old November 20th, 2009, 08:45 AM
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There are two infinite sequences:

(an)n≥1 : 1∙1, 2∙3, 3∙5, 4∙7,……..

(bn)n≥1 : 1∙3, 2∙5, 3∙7, 4∙9,……..

Find general formulae for the two sequences and find how many common elements they have.
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Old November 20th, 2009, 09:21 AM
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Originally Posted by Ilsa View Post
There are two infinite sequences:

(an)n≥1 : 1∙1, 2∙3, 3∙5, 4∙7,……..

(bn)n≥1 : 1∙3, 2∙5, 3∙7, 4∙9,……..

Find general formulae for the two sequences and find how many common elements they have.
a_n=n\cdot(2n-1) and b_n=n\cdot (2n+1)
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Old November 20th, 2009, 10:08 AM
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Hello Ilsa
Quote:
Originally Posted by Ilsa View Post
There are two infinite sequences:

(an)n≥1 : 1∙1, 2∙3, 3∙5, 4∙7,……..

(bn)n≥1 : 1∙3, 2∙5, 3∙7, 4∙9,……..

Find general formulae for the two sequences and find how many common elements they have.
The n^{th} term in sequence a_i is the product of n and the n^{th} odd number. So a_n =n(2n-1).

The n^{th} term in sequence b_i is the product of n and the (n+1)^{th} odd number. So b_n =n(2n+1).

Suppose the n^{th} term in the sequence a_i is equal to the m^{th} term in the sequence b_i. Then
n(2n-1)=m(2m+1)

\Rightarrow 2n^2 -2m^2 = n+m

\Rightarrow 2(n+m)(n-m) = n+m
Now n+m \ne 0, since n \ge1 and m\ge 1. So we can divide both sides by (n+m):
2(n-m)=1

\Rightarrow n-m=\tfrac12
There are no integer solutions to this equation. Hence the two sequences have no terms in common.

Grandad
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