Question #222123

Given the following recurrence relation (M).


an = −4an−1 + 5an−2, a0 = 2, a1 = 8


The solution of (M) is:

a. an = 3 − (−5)

n

b. an = 3 + (5)

n

c. an = (3)

n − 5

d. None of these


Expert's answer

Characteristic equation:

k2+4k−5=0{k^2} + 4k - 5 = 0

D=16+20=36D = 16 + 20 = 36

k1=−4−62=−5{k_1} = \frac{{ - 4 - 6}}{2} = - 5

k2=−4+62=1{k_2} = \frac{{ - 4 + 6}}{2} = 1

Then

an=C1⋅(−5)n+C2⋅1n=C1⋅(−5)n+C2{a_n} = {C_1} \cdot {\left( { - 5} \right)^n} + {C_2} \cdot {1^n} = {C_1} \cdot {\left( { - 5} \right)^n} + {C_2}

a0=2, a1=8⇒{C1+C2=2−5C1+C2=8⇒C1=−1, C2=3{a_0} = 2,\,{a_1} = 8 \Rightarrow \left\{ {\begin{matrix} {{C_1} + {C_2} = 2}\\ { - 5{C_1} + {C_2} = 8} \end{matrix}} \right. \Rightarrow {C_1} = - 1,\,{C_2} = 3

Then

an=−1⋅(−5)n+3=3−(−5)n{a_n} = - 1 \cdot {\left( { - 5} \right)^n} + 3 = 3 - {\left( { - 5} \right)^n}

Answer:a. an=3−(−5)n{a_n} = 3 - {\left( { - 5} \right)^n}


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