PROBLEM 4. Let the sequence $\left(x_{n}\right)_{n \geq 1}$, defined by $x_{1}=3$ and $x_{n}=x_{n-1}+2 n+1$, for $n \geq 2$.

a) Determine the general term of the sequence;

b) Calculate $\lim _{n \rightarrow \infty} n\left(\ln 2+\ln \left(\sum_{k=1}^{n} \frac{1}{x_{2 k-1}}\right)\right)$.[^0]

## NATIONAL MATHEMATICS OLYMPIAD  Local stage - 14.02. 2015  GRADING SCALE - 11th Grade
PROBLEM 4.

$(1 p)$

a) From $x_{n}-x_{n-1}=2 n+1$, for $n \geq 2$

(3p) we obtain that $x_{n}=x_{1}+\sum_{k=2}^{n}\left(x_{k}-x_{k-1}\right)=\sum_{k=1}^{n}(2 k+1)=n(n+2)$

$(2 p)$
b) $\sum_{k=1}^{n} \frac{1}{x_{2 k-1}}=\sum_{k=1}^{n} \frac{1}{(2 k-1)(2 k+1)}=\frac{1}{2} \sum_{k=1}^{n}\left(\frac{1}{2 k-1}-\frac{1}{2 k+1}\right)=$

$=\frac{1}{2}\left(1-\frac{1}{2 n+1}\right)=\frac{n}{2 n+1}$

(1p) Thus, $\lim _{n \rightarrow \infty} n\left(\ln 2+\ln \left(\sum_{k=1}^{n} \frac{1}{x_{2 k-1}}\right)\right)=\ln \left(\lim _{n \rightarrow \infty}\left(\frac{2 n}{2 n+1}\right)^{n}\right)=\ln e^{-\frac{1}{2}}=-\frac{1}{2}$[^1]


[^0]:    ${ }^{1}$ The actual working time is 3 hours;

    ${ }^{2}$ All problems are mandatory;

    ${ }^{3}$ Each problem is graded from 0 to 7.

[^1]:    ${ }^{1}$ Each corrector awards an integer number of points;

    ${ }^{2}$ Any other correct solution is graded accordingly.