## PROBLEM 4.

(2p) It is observed that

| $T_{1}$ | $T_{2}$ | $T_{3}$ | $T_{4}$ | $T_{5}$ | $T_{6}$ | $T_{7}$ | $T_{8}$ | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $2 \cdot 1-1$ | $2 \cdot 2$ | $2 \cdot 3-1$ | $2 \cdot 4$ | $2 \cdot 5-1$ | $2 \cdot 6$ | $2 \cdot 7-1$ | $2 \cdot 8$ | $\ldots$ |

that is, the terms of the sequence are of the form:

- if $n$ is odd, then $T_{n}=2 n-1$
- if $n$ is even, then $T_{n}=2 n$

$(2 p) \quad$ a) The next 2 terms of the sequence are: 21 and 24

(1p) b) The 2015th term is $T_{2015}=2 \cdot 2015-1=4029$

$(2 p)$ c) The number 80 is a term of the sequence, $T_{40}=2 \cdot 40=80$, and the required sum is:

$$
\begin{aligned}
S & =1+4+5+8+\cdots+77+80=(1+5+9+\cdots+77)+(4+8+\cdots+80)= \\
& =78 \cdot 10+4(1+2+3+\cdots+20)=780+2 \cdot 20 \cdot 21=1620
\end{aligned}
$$[^1]


[^0]:    ${ }^{1}$ The actual working time is 2 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.
PROBLEM 4. Consider the sequence with the terms: $1,4,5,8,9,12,13,16,17,20, \ldots$

a) Write the next 2 terms of the sequence;

b) Determine the 2015th term of the sequence;

c) Calculate the sum of the terms less than or equal to 80 from the sequence.[^0]

## NATIONAL MATHEMATICS OLYMPIAD  Local stage - 14.02. 2015 GRADING GUIDE - Grade 5