Sequences advice please- tips on solving nth terms?

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January 27th 2010, 10:40 AM
fishkeeper

Sequences advice please- tips on solving nth terms?

Hi all

We've just started a sequences topic, and Ive never been good at sequences (finding the nth term) at GCSE and now Im struggling to find the nth terms again.

Once Im given the rule its easy, its just finding it

Does anyone have any tips on the best ways to improve myself at finding the nth terms? We have also now started Sigma notation (using r) so any ideas how to help me?

I would appreciate all help, as I really need to brush up my skills, but my text book is utter rubbish at providing advice

Also, I can do simple ones, e.g. 5,9,13,17...

Its ones that involve fractions and negatives for example:

2/3 + 5/9 + 8/27 + 11/81 + 14/243 + 17/729

and

1 x 4 - 3 x 7 + 5 x 10.......+ 29 x46

Help?

Thanks for any advice given!
January 27th 2010, 12:22 PM
running-gag

Quote:

Originally Posted by fishkeeper

Its ones that involve fractions and negatives for example:

2/3 + 5/9 + 8/27 + 11/81 + 14/243 + 17/729

Hi

The denominator is power of 3
The numerator increases by 3 each time
Each fraction is therefore like $\frac{3n-1}{3^n}$

Quote:

Originally Posted by fishkeeper

1 x 4 - 3 x 7 + 5 x 10.......+ 29 x46

The numerator increases by 2 therefore is like 2n+c
The numerator increases by 3 therefore is like 3n+c'
For n=0, 2n+c=1 and 3n+c'=4 lead to c=1 and c'=4
Each fraction is like $\frac{2n+1}{3n+4}$
January 27th 2010, 01:23 PM
Soroban

Hello, fishkeeper!

Quote:

Does anyone have any tips on finding the nth terms?

I can do simple ones: 5, 9, 13, 17, ...

Its ones that involve fractions and negatives . . .

For the sequence: . $\frac{2}{3} + \frac{5}{9} + \frac{8}{27} + \frac{11}{81} + \frac{14}{243} + \frac{17}{729}$

Look at the numerators: . $2,5,8,11,17,\hdots$

Each numerator is 3 more than the preceding numerator.
. . In fact, each numerator is one less than a multiple of 3.

Hence, the general numerator is: . $3n-1$

Look at the denominators: . $3, 9, 27, 81, 243, 729, \hdots$

Each denominator is 3 times the preceding denominator.

Hence, the general denominator is: . $3^n$

The $n^{th}$ term of the series is: . $\frac{3n-1}{3^n}$

Quote:

$(1 \times 4) \:{\color{red}-}\: (3 \times 7) + (5 \times 10) \:{\color{red}-}\: \hdots + (29 \times 46)$
. . . . . ${\color{blue}\uparrow}$
. . . $^{{\color{blue}\text{minus sign?}}}$

If it is an alternating series, we need: . $(-1)^{n-1}$

Look at the first factor of each product: . $1, 3, 5, \hdots, 29$
. . These are the odd numbers: . $(2n-1)$
. . (It stops at the 15th term.)

Look at the second factor of each product: . $4,7,10, \hdots, 46$
. . The second factor is: . $(3n+1)$

The general term is: . $(-1)^{n-1}(2n-1)(3n+1)\quad \text{ for }n \,=\,1,2,3,\hdots, 15$

In sigma notation: . $\sum^{15}_{n=1} (-1)^{n-1}(2n-1)(3n+1)$
January 27th 2010, 10:31 PM
fishkeeper

thanks, but what do you look for when doing sequences to solve them easily? Just generally?