=====6.=====
$\displaystyle T(1)=1$\\
$\displaystyle T(2)=3$\\
$\displaystyle T(n)=T(n-2)+2n-1=T(n-2)+n+(n-1)$ dla $n >2$\\

$\displaystyle T(n)=n+(n-1)+T(n-2)=$\\
$\displaystyle n+(n-1)+(n-2)+(n-3)+T(n-4)=$\\
$\displaystyle n+(n-1)+(n-2)+(n-3)+\cdots+4+3+2+1=$\\
$\displaystyle \frac {n(n+1)} 2$