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 + 3For n odd: U(n) = 5n/2 + 3/2I 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

longtime 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, ....