急 ! F5 sequence #33

ai)
d(n+2)/d(n)=0.8
when n=1,
d(1+2)/d(1)=0.8
d(3)/16=0.8
d(3)=16(0.8)=12.8
when n=3,
d(3+2)/d(3)=0.8
d(5)/12.8=0.8
d(5)=10.24

ii)
d(n+2)/d(n)=0.8
when n=2,
d(2+2)/d(2)=0.8
d(4)/10=0.8
d(4)=8
when n=4,
d(4+2)/d(4)=0.8
d(6)/8=0.8
d(6)=6.4

bi)
by d(n+2)/d(n)=0.8,
when n is a odd number,
d(1) , d(3) , d(5) , ... , d(2n-1) form a geometric sequence with r=0.8.
d(1)+d(3)+d(5)+...+d(2n-1)
=d(1)+d(2x2-1)+d(2x3-1)+...+d(2n-1)
=0.8^(1-1) d(1) + 0.8^(2-1) d(1) + 0.8^(3-1) d(1) + ... + 0.8^(n-1) d(1)
=d(1) + 0.8 d(1) + 0.8^(2) d(1) + ... + 0.8^(n-1) d(1)
=16 + 0.8 x 16 + 0.8^(2) x 16 + ... + 0.8^(n-1) x 16
=16[1 - (0.8)^n]/(1-0.8)
=80[1 - (0.8)^n]
=80 - 80(0.8)^n

ii)
by d(n+2)/d(n)=0.8,
when n is a even number,
d(2) , d(4) , d(6) , ... , d(2n) form a geometric sequence with r=0.8.
d(2)+d(4)+d(6)+...+d(2n)
=d(2)+d(2x2)+d(2x3)+...+d(2n)
=0.8^(1-1) d(2) + 0.8^(2-1) d(2) + 0.8^(3-1) d(2) + ... + 0.8^(n-1) d(2)
=d(1) + 0.8 d(2) + 0.8^(2) d(2) + ... + 0.8^(n-1) d(2)
=10 + 0.8 x 10 + 0.8^(2) x 10 + ... + 0.8^(n-1) x 10
=10[1 - (0.8)^n]/(1-0.8)
=50[1 - (0.8)^n]
=50 - 50(0.8)^n

c)
∵when -1<r<1 , the total distance is a/(1-r)
the total distance of d(1)+d(3)+d(5)+...
=16/(1-0.8)=80
the total distance of d(2)+d(4)+d(6)+...
=10/(1-0.8)=50
∴The total distance=80+50=130
∴The total distance crawled by the ant cannot exceed 130.