Date: Mon, 9 Nov 1998 09:12:58 -0600 (CST) Subject: sequences Name: Ilia
Who is asking: Student
Question: 1) 1, 4, 10, 20, 35, 56, 84,... 2) 0, 6, 30, 90, 210, 420,... Thanks! Hi Ilia One way to approach this type of problem is to take differences of successive terms to form a new sequence, then do the same for those until we come to a sequence of equal differences or to a pattern you recognize and can continue. For example if the given sequence is 0, 2, 8, 21, 44, 80 then this procedure yields a constant row of three 3's. Extending this row to a fourth 3 allows the previous rows to be extended to give 132 at the next term in the original sequence. 0, 2, 8, 21, 44, 80,For your sequence
The same technique will work for the second sequence. Here the first row of differences is a constant times some binomial coefficients.
Cheers
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