Discover the nth Term and Sum of a Series: 12, 23, 60, 169, 494

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The discussion focuses on deriving the nth term and the sum of the series 12, 23, 60, 169, 494. The pattern identified indicates that each term is three times the previous term minus a constant. The nth term is expressed as 3(tn-1) - 3 - 2n. Participants suggest solving the homogeneous recurrence relation and incorporating a linear term with unknown constants to refine the solution. This approach aims to simplify the calculation of tn+1 and the overall sum.

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Homework Statement



Find the nth term and sum to n terms of the series 12,23,60,169,494..

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The Attempt at a Solution


well each term is the triple of the previous term minus some constant.
23=10*3-7
60=23*3-9
..
..
This way i am able to write the nth term as 3(tn-1)-3-2n
and hence proceed to find a telescopic sum by rearrangement and adding.
but this method leaves me with a tn+1 which i am unable to write
So how shud i proceed
 
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hav0c said:
Find the nth term and sum to n terms of the series 12,23,60,169,494..
I guess you meant 10,23, ...
well each term is the triple of the previous term minus some constant.
23=10*3-7
60=23*3-9
..
..
This way i am able to write the nth term as 3(tn-1)-3-2n
and hence proceed to find a telescopic sum by rearrangement and adding.
but this method leaves me with a tn+1 which i am unable to write
So how shud i proceed
Please post your working as far as you got.
Can you solve the homogeneous recurrence relation?
To account for the extra -3-2n, just try adding a linear term with a couple of unknown constants. See if you can determine what those constants would need to be.
 

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