Question

Exercitiul 7 va rog!!!

Answer 1

$\displaystyle \mathtt{a)\sum\limits_{k=1}^n(2k-1)=\sum\limits_{k=1}^n2k-\sum\limits_{k=1}^n1=2\sum\limits_{k=1}^nk-n=\not2 \cdot \frac{n(n+1)}{\not2} -n=}\\ \\ \mathtt{=n^2+n-n=n^2}\\ \\\mathtt{\sum\limits_{k=1}^n(2k-1)=n^2} \\ \\ \mathtt{b)\sum\limits_{k=1}^n(2k+1)=\sum\limits_{k=1}^n2k+\sum\limits{1}=2\sum\limits_{k=1}^nk+n=\not2 \cdot \frac{n(n+1)}{\not 2} +n=}\\ \\ \mathtt{=n^2+n+n=n^2+2n=n(n+2)}\\ \\ \mathtt{\sum\limits_{k=1}^n(2k+1)=n(n+2)}$