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# Sequences - Can't solve this... can you? watch

1. The question is:

Find the nth term of the following sequence:

5, 7, 13, 23, 37
2. (Original post by lalallla22)
The question is:

Find the nth term of the following sequence:

5, 7, 13, 23, 37
The first differences are:

2, 6, 10, 14

There is a constant second difference of 4 which means this is a quadratic sequence. Have you learnt how to find the nth term of a quadratic sequence? If you have, please post what you've tried.
3. [QUOTE=Notnek;74130646]The first differences are:

2, 6, 10, 14

the common difference is 4 which you then half and multiply by n^2

you then right out the 2n^2 sequence as well as the n^2 sequence

2n^2 = 2 8 18 32 50
n^2 = 1 4 9 16 25

The next part I believe I have done incorrectly, however, I shall continue.

I took the 2n^2 sequence away from the original sequence

5 | 7 | 13 | 23 | 37|
- | | | |
2 | 8 | 18 | 32 |50 |
________________

3 | -1| -5 | -9 |-13 |
________________

Then I found the nth term for the sequence above

3, -1, -5, -9, -13
-4 -4 -4 -4

-4n + 7

Then I added it to the first part of the nth term for the initial sequence.

2n^2 - 4n + 7

However, if I then apply that to n:

2* 1^2 - 4*1 + 7

= 2 - 4 +7
= 5

2* 2^2 - 4*2 + 7

= 8 - 8 + 7
= 7

2* 3^2 - 4*3 + 7

=18 - 12 + 7
= 13

Oh, it seems that I have solved it. Never mind.
• the sequence can be written as:5, 5 2,5 2 6 ...
• so it is 5 plus the sum of the sequence 2,6,10,14
• use the sum of geometric series to find how much to add to 5
• sn= n/2(2a (n-1)d)
• n=n-1 as the first tem of the sequence 5,7,13 has none of the other sequence added on
• a=2
• d=4
• therefore the sum of the series is 5 + (n-1)/2 (2(2) (n-2)4)
• which can be simplified to 5 + 2(n-1)^2
4. In case it's any help to anyone, here's an example of finding the nth term of a quadratic sequence, using my favourite method.
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