Školjka

\[S = \sum_{k=0}^{2020}\Bigl\lfloor\frac{2^k}{3}\Bigr\rfloor \] Ispisemo prvih nekoliko elemenata. \[ S = 0 + 0 + 1 + 2 + 5 + 10 + 21 + 42 + 85 + 170 + ... + n_{2020}\] grupiramo \[ S = (0 + 0) + (1 + 2) + (5 + 10) + (21 + 42) + (85 +170) + ... + (n_{2018} + n_{2019}) + n_{2020}\] \[ \iff S =3 + 15 + 63 + 255 + ... + (n_{2018} + n_{2019}) + n_{2020} \] \[ \iff S =(2^2 - 1) + (2^4 - 1) +(2^6 - 1)+(2^8 - 1)+...+ n_{2020} \] \[ \iff S=(\sum_{i=1}^{1009}{4^i -1}) + \frac{4^{1010} - 1}{3} \] primjetimo sumu geometrijskoga niza \[ \iff S = \frac{4^{1010} - 4}{3} + \frac{4^{1010} - 1}{3} - 1009 =\] \[ = \frac{2(4^{1010} -1)}{3} - 1010 \] \[ = \frac{2^{2021} -2 }{3} - 1010 \] \[ 2^{2021} -2 \equiv 2 - 2 \equiv 0 \mod 3 \implies S \in \mathbb{N}\] Rjesenje bi trebalo doraditi. Trebalo bi biti, po mom misljenju, rigoroznije.