Question

Find an explicit formula for the sequence defined recursively by \[a _{1}=1\] and \[a _{k}=2a _{k-1}+6\] Then prove the formula is correct using mathematical induction

Answer 1

First 5 terms (a_1,a_2,a_3,a_4,a_5) are 1,8,22,50,106

Answer 2

That's right.

Answer 3

\[a _{2}=2a _{1}+6\rightarrow2(1)+6\rightarrow2^{1}+6\]

Answer 4

Can you come up with a formula for it?

Answer 5

\[a _{3}=2a _{2}+6\rightarrow2(2+6)+6\rightarrow2^{2}+2*6+6\]

Answer 6

Cant seem to, no

Answer 7

my book says to write them in exponential when you can, which is why i am doing 2^2 instead of 2*2

Answer 8

so I have that form for all the way up to a_5

Answer 9

also, the difference between each term is 1,14,28,56

Answer 10

so i am thinking maybe something to do with 7?

Answer 11

\[a _{5}=2a _{4}+6=2(2^{3}+2^{2}*6+2*6+2*6+6)+6=2^{4}+2^{3}*6+2^{2}*6+2*6+6\]

Answer 12

\[a _{5}=2a _{4}+6=2(2^{3}+2^{2}*6+2*6+2*6+6)+6=2^{4}+2^{3}*6+2^{2}*6+2*6+6\]

Answer 13

Seeing any patterns?

Answer 14

You're doing a brilliant job. Do you notice the pattern with the 2, the 6 and the 7?

Answer 15

We have 2^(n-1) in front everytime...

Answer 16

There's a formula to determine these series when you find out these patters. I forgot what it was called, but assuming that we're starting our count at 1, you can express it as something like\[\frac{a\cdot2^n+b}{c}.\]

Answer 17

Sum of geomretic series?

Answer 18

What is a,b,and c?

Answer 19

Any tips mr.math?

Answer 20

hmm

Answer 21

are there any exponents?

Answer 22

I think so.

Answer 23

I still dont see it for sure im pretty bad at detecting these

Answer 24

obviously the first term is doubling somehow , 1-> 2 -> 4 ->8 -> 16

Answer 25

I'm having a problem myself. I think there could be a better way to do it.

Answer 26

Give me a minute!

Answer 27

Here:
\[a_1=1=1+(2^0-1)(7)\]
\[a_2=1+7=1+(2^1-1)(7)\]
\[a_2=1+3(7)=1+(2^2-1)(7)\]
\[a_3=1+7(7)=1+(2^3-1)(7)\]
...

Answer 28

(2^(n-1)-1)7

Answer 29

I think you can figure it out now. If I find a better method to solving it, I'll let you know.

Answer 30

Yeah, \(\large a(n)=1+7(2^{n-1}-1)\).

Answer 31

Yea, i guess its all just manulation till its in a form where a patern is more recognizable

Answer 32

Have to go now, ill be around later to make sense of this, thanks a lot

Answer 33

It can be simplified into
\[a(n)=7(2^{n-1})-6.\]

Answer 34

You're welcome.

Answer 35

How did you figure this one out?

Answer 36

Was the solution found via just pure math magic manipulation, where you just have to see it. Or was some formula involved or something?

Answer 37

there are techniques to solving these and formulas that one can use. This one is easy enough that you can build up a solution

1
2*1+6=2+6
\[2(2+6)+6=2^2+2\times 6+6=2^2+6(2+1)\]
\[2(2^2+6(2+1))+6=2^3+2\times6(2+1)+6\]\[=2^3+6\times(2^2+2)+6=2^3+6\times(2^2+2+1)\]
...
\[2^{n-1}+6(2^{n-2}+2^{n-2}+\cdots+2^2+2+1)\]
\[=2^{n-1}+6\frac{2^{n-1}-1}{2-1}=2^{n-1}+6(2^{n-1}-1)=2^{n-1}+6\cdot2^{n-1}-6\]
\[=2^{n-1}(1+6)-6=2^{n-1}(7)-6=7\cdot2^{n-1}-6\]

Answer 38

Im trying to figure out how you got from 3rd line up to 2nd line up...

Answer 39

geometric partial sum...
\[\sum_{k=0}^{n-1}a^k=\frac{a^n-1}{a-1}\]

Answer 40

hmm, not familiar

Answer 41

anything to do with this theorem: http://imgur.com/V6Nnz

Answer 42

no....

http://en.wikipedia.org/wiki/Geometric_series#Formula

Answer 43

ok, ill read up on that

Answer 44

But getting to do was a guess right

Answer 45

based on the pattern

Answer 46

Also, do you 2^n-2,2^n-2 on purpose or is it to be 2^n-2,2^n-1

Answer 47

Ah ok, i found that theorem in my book: http://imgur.com/A1g9J

Answer 48

that is a typo ... should be \[2^{n-1}+6(2^{n-2}+2^{n-3}+\cdots+2^2+2+1)\]

Answer 49

should k not be 1 on the bottom of that summation?

Answer 50

we go from 1

Answer 51

k=0...

\[2^0=1\]

Answer 52

We are only given a_1 though

Answer 53

I think I understand it all now, thats everyone, i <3 openstudy