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B. 4019. Prove that


\frac{1}{3^2} + \frac{1}{5^2} + \dots + \frac{1}{{(2n+1)}^2} < \frac{1}{4}.

for every positive integer n.

(Competition problem from the Highlands)

(4 points)

Deadline expired on 15 October 2007.


Google Translation (Sorry, the solution is published in Hungarian only.)

Megoldás: Minden k pozitív egész számra

\Bigl(\frac{1}{2k+1}\Bigr)^2
<\frac{1}{4k^2+4k}=\frac{1}{4}\cdot\frac{1}{k(k+1)}
=\frac{1}{4}\Bigl(\frac{1}{k}-\frac{1}{k+1}\Bigr),

ezért a szóban forgó összeg kisebb, mint

\frac{1}{4}\Bigl\{\Bigl(\frac{1}{1}-\frac{1}{2}\Bigr)+
\Bigl(\frac{1}{2}-\frac{1}{3}\Bigr)+\ldots+
\Bigl(\frac{1}{n}-\frac{1}{n+1}\Bigr)\Bigr\}=
\frac{1}{4}\Bigl(1-\frac{1}{n+1}\Bigr)<\frac{1}{4}.


Statistics on problem B. 4019.
178 students sent a solution.
4 points:133 students.
3 points:8 students.
2 points:5 students.
1 point:6 students.
0 point:26 students.


  • Problems in Mathematics of KöMaL, September 2007

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