Question

A1=1 an+1= an+3 и a1=2 an+1=3an найти а5 рекуррентно

Answer 1

$1)\; \; a_1=1\; ,\; \; a_{n+1}=a_{n}+3\\\\a_2=a_1+3=1+3=4\; ,\\\\a_3=a_2+3=4+3=7\\\\a_4=a_3+3=7+3=10\\\\a_5=a_4+3=10+3=13\\\\ili\\\\a_5=a_4+3=(a_3+3)+3=a_3+2\cdot 3=(a_2+3)+2\cdot 3=\\\\=a_2+3\cdot 3=(a_1+3)+3\cdot 3=a_1+4\cdot 3=1+4\cdot 3=13\\\\\\2)\; \; a_1=2\; ,\; \; a_{n+1}=3\, a_{n}\\\\a_2=3\, a_1=3\cdot 2=6\\\\a_3=3\, a_2=3\cdot 6=18\\\\a_4=3\, a_3=3\cdot 18=54\\\\a_5=3\, a_4=3\cdot 54=162$

$ili\\\\a_5=3\, a_4=3\cdot (3\, a_3)=3^2\cdot a_3=3^2\cdot (3\, a_2)=3^3\cdot a_2=\\\\=3^3\cdot (3\, a_1)=3^4\cdot a_1=3^4\cdot 2=81\cdot 2=162$