how to find the function of the set 1, 6, 31, 156, 781...

Please show your work, thanks

Question

how to find the function of the set 1, 6, 31, 156, 781...

Please show your work, thanks

asnaseer · Answer

see if you can spot a pattern between the differences between each successive term of this sequence.

anonymous · Answer

yes, but HOW to find, for example, the 20th number on the list

asnaseer · Answer

so what pattern did you spot?

anonymous · Answer

previous numer x 5 + 1 = next number

asnaseer · Answer

yes, so you have:$$a_n=5a_{n-1}+1$$with $a_0=0$

asnaseer · Answer

there is a technique to turn this recursive formula into an absolute formula

asnaseer · Answer

I am not an expert in this but it is described here: http://hcmop.wordpress.com/2012/04/20/using-characteristic-equation-to-solve-general-linear-recurrence-relations/ I know @mukushla and @joemath314159 are familiar with this technique

anonymous · Answer

@zscdragon what class is this for?

anonymous · Answer

$$\sum_{n=1}^{\infty} (1+5^n)$$

anonymous · Answer

i am also confused on how you would get the 20th term

asnaseer · Answer

ok, after reading about recurrence relations on the link I gave you above, this is what I found.

we have: $a_n=5a_{n-1}+1$ with $a_0=0$
we first re-arrange this to: $a_n-5a_{n-1}=1$
the solution for this type of recurrence relation is of the form: $a_n=a_{n1}+a_{n2}$

where $a_{n1}$ is the solution to: $a_n-5a_{n-1}=0$ which will be of the form:$$a_{n1}=A5^n$$
and $a_{n2}$ is a polynomial solution of the form defined by the degree of the polynomial on the right-hand-side of $a_n-5a_{n-1}=1$, so in this case we can say:$$a_{n2}=B$$where B is some constant. we then substitute $a_{n2}$ into this equation to get:$$B-5B=1\implies B=-\frac{1}{4}$$

putting these together we get:$$a_n=a_{n1}+a_{n2}=A5^n+B=A5^n-\frac{1}{4}$$and we know $a_0=0$ therefore:$$0=A-\frac{1}{4}\implies A=\frac{1}{4}$$finally we get:$$a_n=\frac{5^n-1}{4}$$this can then be used to find the 20'th term.

I hope I did this correctly. maybe  @mukushla and @joemath314159 can check my work.

more importantly, I hope you understood the process @zscdragon?

anonymous · Answer

@lgbasallote

lgbasallote · Answer

why did you tag me @nickhouraney ?

anonymous · Answer

Looks good to me :)

asnaseer · Answer

@joemath314159 - thanks for confirming :)