A sequence is defined by #a_n=a_(n-1)+2a_(n-2)#, with #a_1=-3#, #a_2=4#. What is the value of #a_6#?

1 Answer
Jan 12, 2017

Answer:

#a_6=14#. The general formulas #a_n=(-1)^n(10/3)+1/6 2^n# gives

#a_10=10/3+1024/6=174#.

Explanation:

Define the shift operator E by

# Ea_(n-2)=a_(n-1)#

Now, the given equation becomes

#f(E)a_n=(E^2-E-2)a_n=0, n = 1, 2, 3, ...#

The general solution of this difference equation is

#a_n=A(rho_1)^n+ B(rho_2)^n#, where #rho_1 and rho_2# are the

zeros of

#f(rho)=rho^2-rho-2=0#, giving #rho = 1 and 2.#

Now,

#a_n=(-1)^nA+2^nB#

The initial value #a_1=-3# gives

#-A+2B=-3#

and #a_2=4# gives

#A+4B=4#.

Solving,

#A = 10/3 and B =1/6#

It follows that

#a_6=10/3+(2^6)(1/6)=14#.