中四數學一題!!!!

Distance travelled by the turtle in n seconds, in m,
= 0.5 + (0.5 + 0.1) + (0.5 + 2 x 0.1) + (0.5 + 3 x 0.1) + ... + [0.5 + 0.1(n-1)]
= 0.5n + 0.1[1 + 2 + 3 + ... + (n-1)]
= 0.5n + 0.1(n)(n - 1) / 2
= n [0.05(n - 1) + 0.5]
Distance travelled by rabbit in n seconds, in m,
= 5 + 5(0.8) + 5(0.8)2 + 5(0.8)3 + ... +5(0.8)n
= 5(1 - 0.8n ) / (1 - 0.8)
= 25(1 - 0.8n )

a) Distance travelled by turtle after 3 seconds
= (2 + 1)(0.05 x 2 + 0.5) m　(Substitute n=2 into the above formula)
= 1.8 m
Distance travelled by rabbit after 3 seconds
= 25(1 - 0.83) m
= 12.2m

Distance between them = (12.2 - 1.8)m = 10.4 m
b) n [0.05(n - 1) + 0.5] = 31.5
0.05n2 + 0.45n - 31.5 = 0
5n2 + 45n - 3150 = 0
n2 + 9n - 630 = 0
(n + 30)(n - 21) = 0
n = -30(rejected) or n = 21

So it takes 21 seconds for turtle to finish the race.

c) The distance travelled by rabbit is 25(1 - 0.8n ) m. The limit of it is 25. Hence rabbit can never finish the race. Turtle will win the race.

RE.a). 問牠們3秒後的距離是多少?
0.5+0.6+0.7 - 5+5*20%+20%(5*20%)
我係F.1的我估係咁~