460 Real Analysis
298.If we were given the recurrence relationxn=xn− 1 +n, for alln, the terms of the
sequence would be the triangular numbersTn=n(n 2 +^1 ). If we were given the recurrence
relationxn=xn− 1 +n−1, the terms of the sequence would beTn− 1 + 1 =n
(^2) −n+ 2
2 .In
our case,
n^2 −n+ 2
2
≤xn≤
n^2 +n
2
.
We expectxn=P (n)/2 for some polynomialP (n)=n^2 +an+b; in fact, we should have
xn=P (n)/ 2 because of the jumps. From here one can easily guess thatxn=n
(^2) + 1
2 ,
and indeed
⌊
n^2 + 1
2
⌋
=
⌊
(n− 1 )^2 + 1
2+
2 (n− 1 )+ 1
2⌋
=
⌊
(n− 1 )^2 + 1
2+
1
2
⌋
+(n− 1 ),which is equal to(n−^1 )
(^2) + 1
2 +(n−^1 )ifnis even, and to
(n− 1 )^2 + 1
2 +nifnis odd.
299.From the hypothesis it follows thata 4 =12,a 5 =25,a 6 =48. We observe that
a 1
1
=
a 2
2= 1 ,
a 3
3= 2 ,
a 4
4= 3 ,
a 5
5= 5 ,
a 6
6= 8
are the first terms of the Fibonacci sequence. We conjecture thatan=nFn, for alln≥1.
This can be proved by induction with the already checked cases as the base case.
The inductive step is
an+ 4 = 2 (n+ 3 )Fn+ 3 +(n+ 2 )Fn+ 2 − 2 (n+ 1 )Fn+ 1 −nFn
= 2 (n+ 3 )Fn+ 3 +(n+ 2 )Fn+ 2 − 2 (n+ 1 )Fn+ 1 −n(Fn+ 2 −Fn+ 1 )
= 2 (n+ 3 )Fn+ 3 + 2 Fn+ 2 −(n+ 2 )(Fn+ 3 −Fn+ 2 )
=(n+ 4 )(Fn+ 3 +Fn+ 2 )=(n+ 4 )Fn+ 4.This proves our claim.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
D. Andrica)
300.The relations
am+am=1
2
(a 2 m+a 0 ) and a 2 m+a 0 =1
2
(a 2 m+a 2 m)implya 2 m= 4 am, as well asa 0 =0. We computea 2 =4,a 4 =16. Also,a 1 +a 3 =
(a 2 +a 4 )/ 2 =10, soa 3 =9. At this point we guess thatak=k^2 for allk≥1.
We prove our guess by induction onk. Suppose thataj =j^2 for allj<k. The
given equation withm=k−1 andn=1 gives