Question

. 322. Найдите шесть первых членов последовательности (а), если: а) а =-5, а, = 1 и ал. = a +а,; , б) а = 2, а, = 4 иа... = а, - а... , + 1 1 - 1 n+ 2 п п+1°​

Answer 1

$\displaystyle\bf\\1)\\a_{1} =-5\\\\a_{2} =1\\\\a_{3} =a_{2+1} =a_{2-1}+a_{2} =a_{1} +a_{2} =-5+1=-4\\\\a_{4} =a_{3+1} =a_{3-1}+a_{3} =a_{2} +a_{3} =1-4=-3\\\\a_{5} =a_{4+1} =a_{4-1}+a_{4} =a_{3} +a_{4} =-4-3=-7\\\\a_{6} =a_{5+1} =a_{5-1}+a_{5} =a_{4} +a_{5} =-3-7=-10\\\\\\2)\\\\a_{1} =2\\\\a_{2} =4\\\\a_{3} =a_{1+2} =a_{1} -a_{1+1}=a_{1}-a_{2} =2-4=-2 \\\\a_{4} =a_{2+2} =a_{2} -a_{2+1}=a_{2}-a_{3} =4-(-2)=4+2=6\\\\a_{5} =a_{3+2} =a_{3} -a_{3+1}=a_{3}-a_{4} =-2-6=-8$

$\displaystyle\bf\\a_{6} =a_{4+2} =a_{4} -a_{4+1}=a_{4}-a_{5} =6-(-8)=6+8=14$