Answer
ek = = (11(4k) – 5)/3
Work Step by Step
ek = 4ek-1 + 5, for all integers k ≥ 1, e0 = 2
e0 = 2
e1 = 4(2) + 5 = 13
e2 = 4(4*2 + 5) + 5 = 57
e3 = 4(4*4*2 + 5*4 + 5) + 5 = 233
e4 = 4(4*4*4*2 + 5*4*4 + 5*4) + 5
= 2*44 + 5*43 + 5*42 + 5*41 + 5*40
= 2*4k + 5*4k-1 + 5*4k-2 + 5*4k-3 + 5*4k-4
= 2*4k + 5(4k-1 + 4k-2 + 4k-3 + 4k-4)
= 2*4k + 5(∑_(i=0)^(k-1)▒4^i )
ek = 2*4k + 5(∑_(i=0)^(k-1)▒4^i )
= 2*4k + 5(4k-1+1 – 1)/(4-1)
= 2*4k + 5(4k – 1)/3
= 2*4k + 5(4k)/3 – 5/3
= 6*4k/3 + 5(4k)/3 – 5/3
= (6(4k) + 5(4k) – 5)/3
= (11(4k) – 5)/3