maths coursework sequencesWatch

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Thread starter 16 years ago
#1
ive got a problem. i have a sequence of 1, 6, 15, 28, 45, 66, 91, 119. i need to find a formula so i can find the nth term and its for my course work so if anybody can help plz let me know ive got 3 days!
thanx
0
16 years ago
#2
Take the first differences between the terms: 5, 9, 13 etc. These are not all equal meaning that the sequence is not linear.

So you then take the second differences, ie the differences between the first differences: 4, 4, 4...

Since these are 4 all the way up, that indicates that the sequence has something to do with 2n^2.

Subtract each term in the sequence from the corresponding value of 2n^2; this will leave you with a second sequence that ought to be linear. Find an expression for this new sequence, put it together with the 2n^2, and hey presto. Instant formula.

Check that it works for a few values of n, and you're done.

Hope this helps.
0
16 years ago
#3
y= 2x^2 -2
0
16 years ago
#4
or rarther nth term = 2n^2 - 2
0
16 years ago
#5
Looks like it should be 2n^2 - n but the last term would need to be 120. Is this possible??? How did you generate the sequence????
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