Problem C. 1618. (September 2020)

C. 1618. Prove that

\(\displaystyle \frac23\le a_{n+1}-a_n<1 \)

holds for all the terms of the sequence \(\displaystyle a_n=\frac{(n-1)n}{n+1}\), with \(\displaystyle n\ge 1\).

(5 pont)

Deadline expired on October 12, 2020.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Mivel

\(\displaystyle a_n=\frac{(n-1)n}{n+1}=\frac{n^2-n}{n+1}=\frac{(n+1)(n-2)+2}{n+1}=n-2+\frac{2}{n+1},\)

ezért

\(\displaystyle a_{n+1}-a_n=n-1+\frac{2}{n+2}-(n-2)-\frac{2}{n+1}=1+\frac{2(n+1)-2(n+2)}{(n+2)(n+1)}=1-\frac{2}{(n+1)(n+2)}.\)

Ha \(\displaystyle n\geq 1\), akkor \(\displaystyle 0<\frac{2}{(n+1)(n+2)}\leq \frac13\), és így

\(\displaystyle \frac23\leq a_{n+1}-a_n<1.\)

Statistics:

267 students sent a solution.
5 points:	115 students.
4 points:	82 students.
3 points:	36 students.
2 points:	13 students.
1 point:	12 students.
0 point:	9 students.

Problems in Mathematics of KöMaL, September 2020