Question

rezolva-ti exercițiul ​

Answer 1

Răspuns:

{0,1}

Explicație pas cu pas:

$(-1)^n = \begin{cases}1, \quad \text{daca } n \in M_2\\-1, \quad \text{daca } n \notin M_2\end{cases}\\\\(2k)^{(2k)} \in M_2\\\\ (2k+1)^{(2k+1)} \notin M_2$

$n\cdot \Bigg (\Big ( (-1)^1+(-1)^2 + \cdots + (-1)^n \Big ) \div \Big ( (-1)^{n^1} + (-1)^{n^2} +\cdots + (-1)^{n^n}\Big ) \Bigg ) = n\cdot \Bigg (\Big ( -1+1-1+1 + \cdots + (-1)^n \Big ) \div \Big ( (-1)^{n^1} + (-1)^{n^2} +\cdots + (-1)^{n^n}\Big ) \Bigg ).\\\\\textrm{Daca }n = 2k\\\\ 2k\cdot \Bigg (\Big ( -1+1-1+1 + \cdots + (-1)^{2k} \Big ) \div \Big ( (-1)^{2k^1} + (-1)^{(2k)^2} +\cdots + (-1)^{(2k)^{2k}}\Big ) \Bigg ) \\\\ = 2k\cdot \Bigg (\Big ( -1+1-1+1 + \cdots + 1 \Big ) \div \Big ( (-1)^{2k} + (-1)^{4k^2} +\cdots + (-1)^{(2k)^{2k}}\Big ) \Bigg ) \\\\= 2k\cdot \Bigg (0 \div \Big ( (-1)^{2k} + (-1)^{4k^2} +\cdots + (-1)^{(2k)^{2k}}\Big ) \Bigg ) = 0\\\\\textrm{Daca } n = 2k+1\\\\= (2k+1)\cdot \Bigg (\Big ( -1+1-1+1 + \cdots + (-1)^{2k+1} \Big ) \div \Big ( (-1)^{(2k+1)^1} + (-1)^{(2k+1)^2} +\cdots + (-1)^{(2k+1)^{(2k+1)}}\Big ) \Bigg ) \\\\ = (2k+1)\cdot \Bigg (\Big ( -1+1-1+1 + \cdots - 1 \Big ) \div \Big ( (-1)^{2k+1} + (-1)^{(4k^2 + 4k +1)} +\cdots + (-1)^{(2k+1)^{(2k+1)}}\Big ) \Bigg )\\\\ = (2k+1)\cdot \Bigg (( - 1 ) \div \Big (-1 -1 -\cdots - (-1)^{(2k+1)^{(2k+1)}}\Big ) \Bigg ) \\\\ = (2k+1)\cdot \Bigg (( - 1 ) \div \Big (-1\cdot (2k+1)\Big ) \Bigg ) = (2k+1)\cdot \Bigg (1 \div (2k+1) \Bigg ) = 1\\\\\textrm{Valorile posibile sunt 0 si 1}$