Chapter 9.4, Problem 47PPSE

Solution

a.

To write an explicit formula for the given sequence.

  $a_n=n+1$

Given:

The given sequence of cans is:

  $2,3,4,5,\dots \dots$

Calculation:

Here, the sequence is an arithmetic where the common difference is $d=1$ and the first term is $a=2$ .

The explicit formula for the sequence as:

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ \Rightarrow a_n=2+\left(n-1\right)1\\ \Rightarrow a_n=2+n-1\\ \Rightarrow a_n=n+1\end{array}$

Hence, the explicit formula is $a_n=n+1.$

b.

To find the summation notation for the series with $10$ cans in the bottom row.

  $S={\displaystyle \sum _{n=1}^9\left(n+1\right)}$

Given:

The sequence is:

  $2,3,4,5,\dots \dots$

Calculation:

The last row contains $10$ cans. Now, finding the term number for $a_n=10$ as:

  $\begin{array}{l}{a}_n=n+1\\ \Rightarrow 10=n+1\\ \Rightarrow n=9\end{array}$

Now, the sum of first $9$ terms of the sequence is:

  $\begin{array}{l}S={\displaystyle \sum _{n=1}^9{a}_n}\\ \Rightarrow S={\displaystyle \sum _{n=1}^9\left(n+1\right)}\end{array}$

c.

To determine the number of cans in ${17}^{th}$ row.

The number of cans in ${17}^{th}$ row is $18.$

Given:

An explicit formula for the given sequence is:

  $a_n=n+1$

Calculation:

Calculating number of cans for ${17}^{th}$ row $(n=17)$ as:

  $\begin{array}{l}{a}_n=n+1\\ \Rightarrow a_{17}=17+1\\ \Rightarrow a_{17}=18\end{array}$

Hence, there would be $18$ cans in the ${17}^{th}$ row.

d.

To explain, could the triangle have $110$ cans or $140$ cans.

No

Given:

An explicit formula for the given sequence is:

  $a_n=n+1$

Calculation:

The first term is:

  $a_1=2$

And the thirteen term is:

  $\begin{array}{l}{a}_3=13+1\\ a_3=14\end{array}$

Now, the sum of first $13$ terms is:

  $\begin{array}{l}{S}_n=\frac{n}{2}\left(a_1+a_n\right)\\ S_{13}=\frac{13}{2}\left(2+14\right)\\ S_{13}=13\times 8\\ S_{13}=104\end{array}$

Next, the fourteen term is:

  $\begin{array}{l}{a}_{14}=14+1\\ a_{14}=15\end{array}$

Now, the sum of first $14$ terms is:

  $\begin{array}{l}{S}_{14}=\frac{14}{2}\left(2+15\right)\\ S_{14}=7\times 17\\ S_{14}=119\end{array}$

Hence,triangle can have $104$ cans or $119$ cans but not $110$ cans.

Next, the fifteen term is:

  $\begin{array}{l}{a}_{15}=15+1\\ a_{15}=16\end{array}$

Now, the sum of first $15$ terms is:

  $\begin{array}{l}{S}_{15}=\frac{15}{2}\left(2+16\right)\\ S_{15}=15\times 9\\ S_{15}=135\end{array}$

Next, the sixteen term is:

  $\begin{array}{l}{a}_{16}=16+1\\ a_{16}=17\end{array}$

And, the sum of first $16$ terms is:

  $\begin{array}{l}{S}_{16}=\frac{16}{2}\left(2+17\right)\\ S_{16}=8\times 19\\ S_{16}=152\end{array}$

Hence, the triangle can have $135$ cans or $152$ cans but not $140$ cans.