## SUBJECT I (7p)

Consider the sequence $3,7,11,15,19$,

a) Complete the sequence with 3 more terms;

b) Determine the 2014th term;

c) Show that the 2014th term is not a perfect square;

d) Calculate the sum of the first 2014 terms.
## Solution:

a) $23,27,31$

$1 p$

b) The general term is of the form $4 k+3$ 2p

the 2014th term is $4 \cdot 2013+3=8055$ $.1 p$
c) $89^{2}<8055<90^{2}$ $1 p$

Or: numbers of the form $4 k+3$ are not perfect squares
d) $S=4 \cdot 0+3+4 \cdot 1+3+4 \cdot 2+3+\ldots \ldots .+4 \cdot 2013+3=4 \cdot(0+1+2+3+\ldots .+2013)+$

$3 \cdot 2014=8114406$ $2 p$