Mamy dane

$a_{n+1}={\left(-1\right)}^n{a}_n+5$  

Oraz

$a_6=4$  

Podstawmy dane do wzoru:

$a_6={\left(-1\right)}^5\cdot a_5+5$

A zatem:

${\left(-1\right)}^5\cdot a_5+5=4\mid -5$  

$-a_5=-1$  

$a_5=1$  

Ze wzoru

${\left(-1\right)}^4\cdot a_4+5=1\mid -5$  

$a_4=-4$  

Ze wzoru

${\left(-1\right)}^3\cdot a_3+5=-4\mid -5$  

$-a_3=-9$  

$a_3=9$  

Ze wzoru

${\left(-1\right)}^2\cdot a_2+5=9\mid -5$  

$a_2=4$  

Ze wzoru

${\left(-a\right)}^1\cdot a_1+5=4\mid -5$  

$-a_1=-1$

$a_1=1$   

Zatem

$\{\begin{array}{l}{a}_1=1\\ a_{n+1}={\left(-1\right)}^n\cdot a_n+5\end{array}$