Chapter 12.7, Problem 45E

a.

To determine

Find the height of third and fourth pile.

Answer to Problem 45E

  $0.02cm,0.04cm$

Explanation of Solution

Given information:

Tear a piece of paper that is $0.005cm$ thick I in half and place two pieces on top of each other, the height of pile of paper is $0.01cm$ . The second pile tear these two pieces of peper in half ,the third pile have four pieces of paper in it.

Calculation:

The height is doubled every time to tear the piece of paper this shows a geometric series where first term is $a_1=0.005$

  $\begin{array}{l}Ratio,r=\frac{a_2}{a_1}=\frac{0.01}{0.005}=2\\ a_3=r\times a_2=2\times 0.01=0.02\\ a_4=r\times a_3=2\times 0.02=0.04\end{array}$

Hence the height of third and fourth pile is $0.02cm,0.04cm$ .

b.

To determine

Write the formula for $n^{th}$ pile.

Answer to Problem 45E

  $a_n=0.005\cdot 2^{n-1}$

Explanation of Solution

Given information:

Tear a piece of paper that is $0.005cm$ thick I in half and place two pieces on top of each other, the height of pile of paper is $0.01cm$ . The second pile tear these two pieces of peper in half ,the third pile have four pieces of paper in it.

Calculation:

The formula is,

  $\begin{array}{l}{a}_n=a_1{r}^{n-1}\\ a_n=0.005\cdot 2^{n-1}\end{array}$

Hence, the formula for $n^{th}$ pile is $a_n=0.005\cdot 2^{n-1}$ .

c.

To determine

Find the height of $10thand100th$ pile.

Answer to Problem 45E

  $\begin{array}{l}{a}_{10}=2.56cm\\ a_{100}\approx 3.169\times {10}^{27}cm\end{array}$

Explanation of Solution

Given information:

Tear a piece of paper that is $0.005cm$ thick I in half and place two pieces on top of each other, the height of pile of paper is $0.01cm$ . The second pile tear these two pieces of peper in half ,the third pile have four pieces of paper in it.

Calculation:

The formula is,

  $\begin{array}{l}{a}_n=0.005\cdot 2^{n-1}\\ a_{10}=0.005\cdot 2^{10-1}\\ a_{10}=0.005\cdot 2^9\\ a_{10}=2.56cm\end{array}$

  $\begin{array}{l}{a}_{100}=0.005\cdot 2^{100-1}\\ a_{100}=0.005\cdot 2^{99}\\ a_{100}\approx 3.169\times {10}^{27}cm\end{array}$

Hence, the height of $10thand100th$ pile

  $\begin{array}{l}{a}_{10}=2.56cm\\ a_{100}\approx 3.169\times {10}^{27}cm\end{array}$