Respuesta :
Answer:
[tex]t_{n}[/tex]=[tex]t_{n-1}[/tex]+[tex]t_{n-2}[/tex] +[tex]t_{n-4}[/tex]
Step-by-step explanation:
[tex]t_{n}[/tex]=multiple ways to climb a tower
When n = 1,
tower= 1 cm
[tex]t_{1}[/tex]= 1
When n = 2,
tower =2 cm
[tex]t_{2}[/tex]= 2
When n = 3,
tower = 3 cm
it can be build if we use three 1 cm blocks
[tex]t_{3}[/tex] = 3
When n = 4,
tower= 4 cm
it can be build if we use four 1 cm blocks
[tex]t_{4}[/tex] = 6
When n > 5
tower height > 4 cm
so we can use 1 cm, 2 cm and 4 cm blocks
so in that case if our last move is 1 cm block then [tex]t_{n-1}[/tex] will be
n —1 cm
if our last move is 2 cm block then [tex]t_{n-2}[/tex] will be
n —2 cm
if our last move is 4 cm block then [tex]t_{n-4}[/tex] will be
n —4 cm
[tex]t_{n}[/tex]=[tex]t_{n-1}[/tex]+[tex]t_{n-2}[/tex] +[tex]t_{n-4}[/tex]
Answer:
tn = t[n-1] + t[n-2] + t[n-4] . . .
Step-by-step explanation:
Let t_n = number of ways of constructing height n.
The last block of a tower of height n can be a 1, 2, or 4.
so it was added to a tower of height n-1, n-2, or n-4 respectively.
Time to try: tn = t[n-1] + t[n-2] + t[n-4] . . . (correct)
1 = 1 from 1
2 = 2 from (2) or (1,1)
3 = 3 from (1,1,1) or (2,1) or (1,2)
4 = 6 from (1,1,1,1) (1,1,2) (1,2,1) (2,1,1) (2,2) (4)
5 = 10 from
1 x (1,1,1,1,1)
4 x (1,1,1,2)
3 x (1,2,2)
2 x (1,4)
6 = 18 from
2 x (4,2) ...
3 x (4,1,1) ...
1 x (2,2,2) ...
6 x (2,2,1,1) ...
5 x (2,1,1,1,1) ..
1 x (1,1,1,1,1,1) ..
Prediction for 7 is t6 + t5 + t3 = 18 + 10 + 3 = 31
6 x (4,2,1)
4 x (4,1,1,1)
4 x (2,2,2,1)
10 x (2,2,1,1,1)
6 x (2,1,1,1,1,1)
1 x (1,1,1,1,1,1,1)
Total =31
