8. Zadatak

Promotrimo zbroj svih danih brojeva:

$$S=\binom{2n+1}{1}+\binom{2n+1}{2}+\cdots+\binom{2n+1}{n}.$$

Kako je

$$\binom{2n+1}{k}=\binom{2n+1}{2n+1-k},$$

za $k\in\{1,2,\ldots,n\}$, vrijedi

$$\begin{aligned} S&=\binom{2n+1}{1}+\binom{2n+1}{2}+\cdots+\binom{2n+1}{n}\\ &=\binom{2n+1}{2n}+\binom{2n+1}{2n-1}+\cdots+\binom{2n+1}{n+1}. \end{aligned}$$

Prema binomnom poučku,

$$\begin{aligned} (1+1)^{2n+1} &=\binom{2n+1}{0}+\binom{2n+1}{1}+\cdots+\binom{2n+1}{2n}+\binom{2n+1}{2n+1}\\ &=1+S+S+1. \end{aligned}$$

pa je

$$2S+2=2^{2n+1},$$

odnosno

$$S=2^{2n}-1.$$

Dakle, zbroj svih danih brojeva je neparan broj. Iz toga zaključujemo da među danim brojevima ima neparno mnogo neparnih brojeva.