Question

Gretchen has a set of blocks of heights 1, 2, and 4-centimeters. Imagine stacking the blocks one on top of the other to make a tower of height n centimeters. For example, a tower of height 6 centimeters could be made by stacking 6 1-cm blocks, 3 2-cm blocks, a 4-cm block on top of a 2- cm block, a 2-cm block on top of a 4-cm block, etc. (Order in the stack matters, think of the blocks of different heights as different colors and the stack looks different if two different color blocks are switched.) Let me be the number of ways to construct a tower of height n centimeters. Find a recursive formula for nB and try to find a formula that only depends on n.

Answer 1

Answer:

$t_{n}$=$t_{n-1}$+$t_{n-2}$ +$t_{n-4}$

Step-by-step explanation:

$t_{n}$=multiple ways to climb a tower

When n = 1,

tower= 1 cm  

$t_{1}$= 1

When n = 2,

tower =2 cm   

$t_{2}$= 2

When n = 3,

tower = 3 cm

it can be build if we use three 1 cm blocks

$t_{3}$ = 3

When n = 4,

tower= 4 cm

it can be build if we use four 1 cm blocks

$t_{4}$ = 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 $t_{n-1}$ will be

n —1 cm

if our last move is 2 cm block then $t_{n-2}$ will be

n —2 cm

if our last move is 4 cm block then $t_{n-4}$ will be

n —4 cm

 

$t_{n}$=$t_{n-1}$+$t_{n-2}$ +$t_{n-4}$

Answer 2

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