## find out nth term of sequence 5, 13, 25, 41, 61

Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc.
confused
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Joined: Sun Feb 22, 2009 10:07 pm

### find out nth term of sequence 5, 13, 25, 41, 61

find out the nth term of the following sequences:

5, 13, 25, 41, 61

also tell me how to work out the nth term of this sequence

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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find out the nth term of the following sequences: 5, 13, 25, 41, 61
A good first step is often to see what the differences and ratios are. But since 13 is prime and 25 isn't a multiple, this is probably not a purely multiplicative sequence. Also, I don't "see" any obvious evidence of squaring or cubing (such as numbers related to 4, 9, 16, 25,... or 8, 27, 64,...).

The differences, though, look promising: 13 - 5 = 8, 25 - 13 = 12, 41 - 25 = 16, 61 - 41 = 20. For each of these, the difference was a multiple of four. In fact, the first difference was twice four, the second difference was three times four, and so forth.

So see if you can come up with a formula for this difference, in terms of the n-th term.

Then see if you can develop a formula for the entire n-th term, both the added multiple of four and the original value.
also tell me how to work out the nth term of this sequence
For some general methods and ideas, try here.

confused
Posts: 20
Joined: Sun Feb 22, 2009 10:07 pm

### Re: find out nth term of sequence 5, 13, 25, 41, 61

is this what i should do?
5, 13, 25, 41, 61
5, 5+8, 5+8+12, 5+8+12+16, 5+8+12+16+20
then what?

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
5, 13, 25, 41, 61
5, 5+8, 5+8+12, 5+8+12+16, 5+8+12+16+20
then what?
Now you need to find a way of relating the above pattern to the "n" which stands for the "n-th" place.

The first entry is just 5.

The second entry is 5 + 2(4).

The third entry is 5 + 2(4) + 3(4) = 5 + 5(4).

The fourth entry is 5 + 2(4) + 3(4) + 4(4) = 5 + 9(4).

The fifth entry is 5 + 2(4) + 3(4) + 4(4) + 5(4) = 5 + 14(4).

The multipliers of 4, at each stage, are 2, 5, 9, and 14. Note that these grow by amounts that increase, at each stage, by 1. Can we use this at all, in relating the pattern to "n"?

The first multiplier, when n = 1, is 0.

The second multiplier, when n = 2, is 2.

The third multiplier, when n = 3, is 5.

The fourth multipler, when n = 4, is 9.

The fifth multipler, when n = 5, is 14.

From what you've learned about the method of common differences, you know how to determine that the pattern for these multipliers in terms of "n".

Then plug this pattern into the larger pattern: 5 + (pattern for the n-th term's multipler)(4).

Gotta run! I'll try to respond more later, if you like.

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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Back now...

Anyway, using the techniques in the lesson in the link provided earlier, you'll have already proven to yourself that the multipliers follow some quadratic rule. For the first three terms (and thus the first three values of "n'), the multipliers are then given by:

. . . . .$n:\, an^2\, +\, bn\, +\, c\, =\, \mbox{(multiplier)}$
. . . . .$1:\, a\, +\, b\, +\, c\, =\, 0$
. . . . .$2:\, 4a\, +\, 2b\, +\, c\, =\, 2$
. . . . .$3:\, 9a\, +\, 3b\, +\, c\, =\, 5$

Solving this system of equations for the coefficients, I believe you end up with:

. . . . .$\mbox{(multiplier) }\, =\, \frac{n^2\, +\, 3n\, -\, 4}{3}$

Can you take it from there?

confused
Posts: 20
Joined: Sun Feb 22, 2009 10:07 pm

ok