Question

Find the first five terms of the following sequence. \[ a_{0}=1, a_{1}=2, a_{2}=\frac{a_{1}-1}{a_{0}-3}+3, n=2, n \in N \]

Answer 1

\(\because\) a₂ = \(\frac{a_1-1}{a_0-3}+3=\frac{a_{2-1}}{a_{2-2}-3}+3\)

\(\therefore\) a_n = \(\frac{a_{n-1}-1}{a_{n-2}-3}+3\)

\(\therefore\) a₂ = \(\frac{a_1-1}{a_0-3}+3=\frac{2-1}{1-3}+3\) = \(\frac{1}{-2}+3=\frac52\)

a₃ = \(\frac{a_{2}-1}{a_1-3}+3\) = \(\frac{5/2-1}{2-3}+3\) = \(\frac{3/2}{-1}+3=3-\frac32=\frac32\)

a₄ = \(\frac{a_{3}-1}{a_2-3}+3\) = \(\frac{3/2-1}{5/2-3}+3\) = \(\frac{1/2}{-1/2}+3\) = 3 - 1 = 2

a₅ = \(\frac{a_{4}-1}{a_3-3}+3\) = \(\cfrac{2-1}{\frac32-3}+3\) = \(\frac{1}{-3/2}+3\) = \(\frac{-2}3+3\) 

 = \(\frac{9-2}3 = 7/3\)

Hence, first five terms of given sequence is a₀ = 1, a₁ = 2, a₂ = 5/2, a₃ = 3/2, a₄ = 2