Let U(n) be the sequence with n a natural number such that :
U(2n+1)-U(2n)=1
U(2n)-U(2n-1)=4
U(0)=3
Find the explicit formula of this sequence, if this is possible.
I honestly don't know how to begin.
Except that it's fairly easy to see that:
- when n is even, U(n) = U(n-1)+4
- when n is odd, U(n)=U(n-1)+1
For n even: U(n) = 5n/2 + 3
For n odd: U(n) = 5n/2 + 3/2
I just wrote out the first 8 terms
3,4,8,9,13,14,18,19,... and separated the odd and even positions
3,8,13,18,..
4,9,13,19,...
Both are arithmetic with common difference = 5 and I just picked the constant to start in the right place.
You can also add your 2 recursion formulas to get: U(2n+1) - U(2n-1) = 5 so
U(2n+1) = 5 + U(2n-1). i.e. go back 2 and add 5 and that helped me get the 5n/2
Oh i didn't think in that way thanks !
Sorry for the small size but you can blow it up by clicking the image. It took a long time but I finally got my head in the right place. I tried to use Generating Functions , which works for Fibonacci, but not so well here. At any rate, I got it !!
I tried to explain as I went along but feel free to ask more questions. What a great problem - I learned a lot about how to apply alternating sequences along the way. Thanks very much.
BPRP are you out there?? IMHO this definitely deserves a video.
Good use of the (-1)^n. I didn't see it ! It's easier than using cos !
I made a slip in (I): s(n) = (-1)^n should read 1,-1,1,-1,1,-1,.... not 1,0,1,0,1,0, ....