Recurrence Relations, calculating

Question

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How do you calculate u2 and u3? u2 = 2(2)-3, u3 = 2(3)-3?

Answer 1

i) $u_0=4$, $u_{n+1}= 2u_n -3$. \begin{align} \mbox{for} : n=0,\, u_{1}&= 2u_0 -3 =2\times 4 -3 = 8-3=5; \\ \mbox{for} : n=1,\,u_{2}&= 2u_1 -3=2\times 5 -3 =7 ;\\ \mbox{for} : n=2,\, u_{3}&= 2u_2 -3 =2\times 7-3 =11 ;\\ \mbox{for} : n=3,\, u_{4}&= 2u_3 -3= .... \end{align} Trying to doing the same thing for ii).

Answer 2

i. u0 = 4, un+1 = 2un – 3

for n=0, u1=2u0 -3 = 2(4) – 3 = 8-3 = 5

for n=1, u2=2u1 -3 = 2(5) – 3 = 10-3 = 7

for n=2, u3=2u2 -3 = 2(7) -3 = 14-3 = 11

for n=3, u4=2u3 – 3 = 2(11)- 3 =22-3 = 19

ii. u0 = 81, un+1 = (-1/3)un

for n=0, u1 = (-1/3)u0=(-1/3)(81) = -27

for n=1, u2 = (-1/3)u1=(-1/3)(-27) = 9

for n=2, u3 = (-1/3)u2=(-1/3)(9)=-3

for n=3, u4 = (-1/3)u3=(-1/3)(-3)=1