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\showsolutionstrue
\begin{document}
\solset{5}{March 23, 2001}

\textbf{Be sure to indicate your TA's name clearly on the solutions
you turn in}.

\paragraph{Reading}
Rosen, \textit{Discrete Mathematics and Its Applications},
Sections 1.7 and~1.8.


\paragraph{Policy on collaboration}
If you collaborated with someone on the problem sets, you should
mention their names on top of the problem set. If you did not
collaborate with anyone, you should explicitly say so. As a reminder,
the write-up of your solutions should be entirely your own.

\paragraph{Self-study}

The following exercises are \emph{optional;} do them if you think you
need the practice. The mini-quiz in tutorial on Friday will draw
(only) from these problems.

Exercises from Rosen (answers in the back of the book): Section 1.7:
5, 8, 9.  Section 1.8: 7, 13, 14, 21


\medskip
\begin{problems} 
%Annuities
%pset5 problem1 spring98
\problem {\bf (10 points)}

A college loan officer offers prospective M.Eng students the following
deal.  He will give them \$23,100 at the start of each year for five years
starting today to pay for their college tuition.  He offers the following
plan for repayment:\\
Wait five years, then repay \$$x.00$ at the start
of each year for the next ten years.

Suppose the annual interest rate is 8\%.  What value of $x$ will make the 
present value of the repayment equal to the present value of the loan?

\solution { \textbf{Solution:}\\
The formula for finding the present value of an annuity (given in lecture)
is
\[V = m \sum_{j=0}^{n-1}{1 \over (1+p)^j}\]
where $V$ is the present value, $m$ is the dollar amount of each payment, $n$ is
the number of years of the annuity, and $p = 0.08$ is the current
interest rate.

So the present value of what the college loan officer offers is
\begin{eqnarray*}
V & = & 23100 \cdot {\displaystyle\sum_{j=0}^4(\frac{1}{1+p})^j} \\
 & = & 23100 \cdot {\displaystyle\frac {(1 - \frac{1}{(1 + p)^5})} {(1 -
 \frac{1}{1 + p})}} \\
 & = & 23100 \cdot 4.312\\
 & = & \$99,610.13.
\end{eqnarray*}

The value, $V'$, of the repayment is
\[V' = x \cdot \sum_{j=5}^{14} \frac{1}{(1+p)^j}.\]

Since we want the present values to be equal, let $V' = V$, and solve for
$x$:
\begin{eqnarray*}
x & = & {\$99,610.13 \over { (\frac{1}{1+p})^5} \cdot (\sum_{j=0}^9(\frac{1}{1+p})^j) }\\
 & = & \frac {\$99,610.13} { (\frac{1} {1+p})^5 \cdot { {(1 - \frac{1}{(1+p)^{10}})} \over {(1 - \frac{1}{1+p})}}}\\
 & = & \frac  {\$99,610.13}{0.6805 \cdot 7.246}\\
 & = & \$20,196.25.
\end{eqnarray*}

So, rounding off to the nearest dollar, $x=\$20, 196$.
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%summations
%S99 PS 5

\problem {\bf (10 points)} Find closed-form expressions for the
following sums.

\begin{problemparts}

\iffalse

%\problempart
\[
\sum_{i=0}^n \frac{9^n - 7^n}{11^n}
\]

\solution {\textbf{Solution:}\\
Split the expression into two geometric series and then
apply the formula for the sum of a geometric series.

\begin{eqnarray*}
\sum_{i=0}^n \frac{9^i - 7^i}{11^i}
    & = &   \sum_{i=0}^n \left(\frac{9}{11}\right)^i -
        \sum_{i=0}^n \left(\frac{7}{11}\right)^i \\
    & = &   \frac{1 - \left(\frac{9}{11}\right)^{n+1}}{1 - \frac{9}{11}}
          - \frac{1 - \left(\frac{7}{11}\right)^{n+1}}{1 - \frac{7}{11}} \\
    & = &   - \frac{11}{2} \cdot \left(\frac{9}{11}\right)^{n+1}
        + \frac{11}{4} \cdot \left(\frac{7}{11}\right)^{n+1}
        + \frac{11}{4}
\end{eqnarray*}

}
\fi

\problempart {\bf (2.5 points)}
\[
\sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} 2^{-(i +2j +3k)}
\]

\solution {\textbf{Solution:}\\
 Apply the formula for the sum of a geometric series three
times.

\begin{eqnarray*}
\sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} 2^{-(i +2j +3k)}
    & = &   \sum_{i=0}^{\infty} 2^{-i} \cdot
        \left(\sum_{j=0}^{\infty} 2^{-2j} \cdot
        \left(\sum_{k=0}^{\infty} 2^{-3k} \right) \right) \\
    & = &   \sum_{i=0}^{\infty} 2^{-i} \cdot
        \left(\sum_{j=0}^{\infty} 2^{-2j} \cdot
        \frac{8}{7} \right) \\
    & = &   \sum_{i=0}^{\infty} 2^{-i} \cdot
        \frac{4}{3} \cdot \frac{8}{7} \\
    & = &   \frac{2}{1} \cdot \frac{4}{3} \cdot \frac{8}{7} \\
    & = &   \frac{64}{21}
\end{eqnarray*}
}

%\iffalse

\problempart {\bf (2.5 points)}
\[
\sum_{i=n+1}^{2n} i^2
\]

\solution {\textbf{Solution:}\\

This expression is the difference of two sums of squares, which can be
reduced to a closed form by a standard formula.

\begin{eqnarray*}
\sum_{i=n+1}^{2n} i^2
    & = &   \sum_{i=1}^{2n} i^2 - \sum_{i=1}^{n} i^2 \\
    & = &   \frac{(2n)(2n+\frac{1}{2})(2n+1)}{3} -
        \frac{n(n+\frac{1}{2})(n+1)}{3} \\
    & = &   \frac{7}{3} n^3 + \frac{3}{2} n^2 + \frac{1}{6} n \\
    & = &   \frac{n (2n + 1) (7n + 1)}{6}
\end{eqnarray*}
}

\problempart {\bf (2.5 points)}
\[
\prod_{i=1}^{n} 3^{4i + 5}
\]

\solution {
\textbf{Solution:}\\
This product can be reduced to an easy sum by taking a
logarithm.

\begin{eqnarray*}
\prod_{i=1}^{n} 3^{4i + 5}
    & = &   3^{\log_3\left( \prod_{i=1}^{n} 3^{4i + 5} \right)} \\
    & = &   3^{\sum_{i=1}^n 4i+5} \\
    & = &   3^{2n(n+1) + 5n}
\end{eqnarray*}
}
%\fi

\problempart {\bf (2.5 points)}

\[
\sum_{j=1}^n \sum_{i=0}^{\infty} j^{2/3} \cdot \left(1 - \frac{1}{2 j^{1/3}}\right)^i
\]

\solution {
\textbf{Solution:}\\
This fearsome-looking sum is a paper tiger; we just apply
the formula for the sum of a geometric series followed by the formula
for the sum of an arithmetic series.

\begin{eqnarray*}
\sum_{j=1}^n \sum_{i=0}^{\infty} j^{2/3} \cdot \left(1 - \frac{1}{2 j^{1/3}}\right)^i
    & = &   \sum_{j=1}^n j^{2/3} \cdot \frac{1}{1 - \left(1 - \frac{1}{2 j^{1/3}}\right)} \\
    & = &   \sum_{j=1}^n 2j \\
    & = &   n (n+1)
\end{eqnarray*}
}
\iffalse

\problempart

\[
\prod_{i=2}^{n} \left(1 - \frac{1}{i^2}\right)
\]

\solution { 
\textbf{Solution:}\\
We give two solutions.  First, evaluating a few terms
suggests that the product is equal to

\[
\frac{n+1}{2n}.
\]

We could verify this by induction.  The calculation needed for the
inductive step is shown below.

\begin{eqnarray*}
\prod_{i=2}^{n+1} \left(1 - \frac{1}{i^2}\right)
    & = &   \left(1 - \frac{1}{(n+1)^2}\right) \cdot
        \prod_{i=2}^{n} \left(1 - \frac{1}{i^2}\right) \\
    & = &   \left(1 - \frac{1}{(n+1)^2}\right) \cdot
        \frac{n+1}{2n} \\
    & = &   \frac{(n+1)+1}{2(n+1)}
\end{eqnarray*}

As an alternative approach, we might observe that this is a
``telescoping product''.

\begin{eqnarray*}
\prod_{i=2}^{n} \left(1 - \frac{1}{i^2}\right)
    & = & \prod_{i=2}^{n} \frac{(i-1)(i+1)}{i^2} \\
    & = & \frac{1 \cdot 3}{2 \cdot 2}
        \cdot \frac{2 \cdot 4}{3 \cdot 3}
        \cdot \frac{3 \cdot 5}{4 \cdot 4} \cdots
        \frac{(n-1) \cdot (n+1)}{n \cdot n} \\
    & = &   \frac{1}{2} \cdot {n+1}{n}
\end{eqnarray*}
}
\fi

\end{problemparts}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%integral method, Fall 98 PS5
\problem {\bf (10 points)} For the following summation use the integral method to find
upper and lower bounds that differ by at most 0.1.  Note that you may
need to add the first few terms explicitly and then use integrals to
bound the sum of the remaining terms.

\[
\sum_{i=1}^{\infty} \frac{1}{i^2}
\]

\solution { \textbf{Solution:}\\

We can upper bound the summation as follows:

\begin{eqnarray*}
\sum_{i=1}^{\infty} \frac{1}{i^2}
    & \leq &    \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \int_3^{\infty} \frac{1}{x^2} \ dx \\
    & = &       \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \left( - \frac{1}{x} \right)_3^{\infty} \\
    & = &       \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \frac{1}{3} \\
    & = &       1.694\ldots
\end{eqnarray*}

We can lower bound the summation similarly:

\begin{eqnarray*}
\sum_{i=1}^{\infty} \frac{1}{i^2}
    & \geq &    \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \int_3^{\infty} \frac{1}{(x+1)^2} \ dx \\
    & = &       \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \left( - \frac{1}{x+1} \right)_3^{\infty} \\
    & = &       \frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
            \frac{1}{4} \\
    & = &       1.611\ldots
\end{eqnarray*}
}

(The actual value of the summation turns out to be $\pi^2 / 6 =
1.644\ldots$.)  


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%pset5 spring98
%using derivatives to figure out sums
\problem {\bf(10 points)}
\begin{problemparts}
\problempart {\bf (5 points)} Find a closed form solution for
$\sum_{i=1}^n i^2x^i$.

\solution {\textbf{Solution:}\\
From lecture we know that
\[{\displaystyle \sum_{i=1}^n ix^i} = \frac{x - (n+1)x^{n+1} + nx^{n+2}}{(1-x)^2}.\] 
Differentiating (with respect to $x$) the above expression and
multiplying with $x$ we find that 
\begin{eqnarray*}
{\displaystyle \sum_{i=1}^n i^2x^{i-1}} &=& \frac{1+x-(n+1)^2x^{n}+(2n^2+2n-1)x^{n+1}-n^2x^{n+2}}{(1-x)^3}\\
{\displaystyle \sum_{i=1}^n i^2x^i} &=& \frac{x+x^2-(n+1)^2x^{n+1}+(2n^2+2n-1)x^{n+2}-n^2x^{n+3}}{(1-x)^3}
\end{eqnarray*}
}

\problempart {\bf (5 points)} Verify your closed form by induction.

\solution {\textbf{Solution:}\\
Proof by induction. Let $P(n)$ be the statement that 
\[{\displaystyle \sum_{i=1}^n i^2x^i} = \frac{x+x^2-(n+1)^2x^{n+1}+(2n^2+2n-1)x^{n+2}-n^2x^{n+3}}{(1-x)^3}.\]
Base Case. $P(1)$ is true since
\[{\displaystyle \sum_{i=1}^1 i^2x^i} = x = \frac{x+x^2-4x^2+3x^3-x^4}{(1-x)^3}.\]
Inductive Step. Assume $P(n)$ is true. Then
\begin{eqnarray*}
{\displaystyle \sum_{i=1}^{n+1} i^2x^i}
& = & {\displaystyle \sum_{i=1}^n i^2x^i} + (n+1)^2x^{n+1} \\
 & = & \frac{x+x^2-(n+1)^2x^{n+1}+(2n^2+2n-1)x^{n+2}-n^2x^{n+3}}{(1-x)^3} + (n+1)^2x^{n+1} \\
& = & \frac{x+x^2-(n+2)^2x^{n+2}+(2(n+1)^2+2(n+1)-1)x^{n+3}-(n+1)^2x^{n+4}}{(1-x)^3}
\end{eqnarray*}
Thus $P(n+1)$ is true. Hence, $P(n)$ is true for all $n$.
}
\end{problemparts}
              


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%fall 1998 PS 5
%solving recurrences by plugging and chugging, verifying using induction
\problem {\bf (10 points)} Consider the following recurrence.
\begin{eqnarray*}
y_0 & = & 1 \\ y_n & = & \sum_{i=1}^n y_{n-i} \cdot 2^i
\end{eqnarray*}

\begin{problemparts}

\problempart {\bf (5 points)} Guess the solution to the recurrence. \hint Look at the first few terms

\solution { \textbf{Solution:}\\

First, we compute the first few terms to form the basis for a guess.

\begin{eqnarray*}
y_0 & = &   1 \\
y_1 & = &   2 \cdot y_0
     =  2 \cdot 1
     =  2 \\
y_2 & = &   2 \cdot y_1 + 4 \cdot y_0
     =  2 \cdot 2 + 4 \cdot 1
     =  8 \\
y_3 & = &   2 \cdot y_2 + 4 \cdot y_1 + 8 \cdot y_0
     =  2 \cdot 8 + 4 \cdot 2 + 8 \cdot 1
     =  32 \\
y_4 & = &   2 \cdot y_3 + 4 \cdot y_2 + 8 \cdot y_1 + 16 \cdot y_0
     =  2 \cdot 32 + 4 \cdot 8 + 8 \cdot 2 + 16 \cdot 1
     =  128 \\
\end{eqnarray*}

The above calculation suggests the following claim.

\medskip
{\bf Claim.}
The solution to the recurrence is $y_0 = 1$ and $y_n = 2^{2n-1}$ for $n
\geq 1$.
\medskip
}

\problempart {\bf (5 points)} Prove your guess correct using strong induction.

\solution { \textbf{Solution:}\\

\begin{proof}
The proof is by strong induction.  Let $P(n)$ be the proposition that
$y_0 = 1$ and $y_n = 2^{2n-1}$ for $n \geq 1$.  In the base case,
$P(0)$ holds trivially.  In the inductive step, assume $P(0), \ldots,
P(n)$ to prove $P(n+1)$.  By the definition of $y_{n+1}$, we have:

\begin{eqnarray*}
y_{n+1} & = &   \sum_{i=1}^{n+1} y_{n+1-i} \cdot 2^i \\
    & = &   y_0 \cdot 2^{n+1} + \sum_{i=1}^n y_{n+1-i} \cdot 2^i \\
    & = &   1 \cdot 2^{n+1} + \sum_{i=1}^n 2^{2(n+1-i)-1} \cdot 2^i \\
    & = &   2^{n+1} + 2^{2n+1} \cdot \sum_{i=1}^{n} \frac{1}{2^i} \\
    & = &   2^{n+1} + 2^{2n+1} \cdot \left( 1 - \frac{1}{2^n} \right) \\
    & = &   2^{2(n+1) - 1}
\end{eqnarray*}

In the second step, we split out the special term of the summation
that contains $y_0$.  The third step uses the induction hypotheses,
and in the fourth step we pull a constant out of the summation.  The
fifth equation follows from the formula for the sum of a geometric
series.  The last step is simplication.

This shows that $P(0), \ldots, P(n)$ imply $P(n+1)$, and the claim is
proved by strong induction.
\end{proof}
}

\end{problemparts}   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%fall 1998 PS 5
%coming up with recurrences,
%solving them by plugging and chugging, verifying using induction 
\problem {\bf (10 points)}  Suppose that $n$ ovals are drawn in a plane
so that every pair of ovals intersects at either 0 or 2 points.
Let $x_n$ be the maximum number of regions that can be formed by such
a configuration of ovals.  We can express $x_n$ using the
recurrence below.

\begin{eqnarray*}
x_1 & = & 2 \\
x_{n+1} & = & x_n + 2n
\end{eqnarray*}

The region outside of any oval ``counts''; for example, one oval
defines two regions, because the oval has an inside region and an
outside region.

\begin{problemparts}

\problempart {\bf (7 points)} Solve this recurrence using the plug and chug method.

\solution {\textbf{Solution:}\\
First, we repeatedly apply the recurrence equation.

\begin{eqnarray*}
x_n & = & x_{n-1} + 2(n-1) \\
    & = & x_{n-2} + 2(n-2) + 2(n-1) \\
    & = & x_{n-3} + 2(n-3) + 2(n-2) + 2(n-1)
\end{eqnarray*}

We can now guess a general form and check it by applying the
recurrence equation one more time.

\begin{eqnarray*}
x_n & = &   x_{n-i} + 2(n-i) + \ldots + 2(n-2) + 2(n-1) \\
x_n
  & = & x_{n-i-1} + 2(n-i-1) + 2(n-i) + \ldots + 2(n-2) + 2(n-1) \\
  & = & x_{n-(i+1)} + 2(n-(i+1)) + 2(n-i) + \ldots + 2(n-2) + 2(n-1)
\end{eqnarray*}

Setting $i = n-1$, gives an expression for $x_n$.

\begin{eqnarray*}
x_n & = &   x_{n-i} + 2(n-i) + \ldots + 2(n-2) + 2(n-1) \\
    & = &   x_1 + 2 + 4 + \ldots + 2(n-2) + 2(n-1) \\
    & = &   2 + 2 + 4 + \ldots + 2(n-2) + 2(n-1) \\
    & = &   2 + 2 \sum_{j=1}^{n-1} j
\end{eqnarray*}

Now we get the closed form $x_n = n(n-1) + 2$ from the formula for the
sum of an arithmetic series.
}

\problempart {\bf (3 points)} Describe how to arrange $n$ ovals so as to form $x_n$ regions.
No proof of correctness is required.

\solution {
\textbf{Solution:}\\
To do this, each new oval should intersect each previous oval in two
places as well as have intersect the innermost region (contained by
all the previous ovals).  One way to do this is to make all the ovals
circles of the same size and place the second circle slightly right of
the first, the third slightly right of the second, and so forth.
}

\end{problemparts}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problem {\bf (10 points)} Shady finance people often assert that a loan can be paid off
more quickly by making half-size payments twice as often.  The theory
is that frequent payments prevent interest from building up.  Suppose
that I take out a loan for $L$ dollars, and at the end of every month,
the bank charges me interest on the loan at rate $i$.

\begin{problemparts}

\problempart First, let's consider a ``frequent payment'' repayment
strategy.  Suppose that I pay $P$ dollars to the bank every month,
right after the bank charges me interest.  (So, after 1 month I owe $L
(1 + i) - P$ dollars, and after 2 months I owe $L (1 + i)^2 - P \cdot
(1 + i) - P$ dollars.)  Give a closed-form expression for the number
of months until the loan is repaid.

\bigskip

(Assume that interest charges are smaller than my monthly payment;
that is, $i L < P$.  Do not worry about rounding to an integral number
of months.)

\solution{
\textbf{Solution:}\\
After $k$ months, I owe

\begin{eqnarray*}
\lefteqn{L (1+i)^k - P \cdot (1+i)^{k-1} - P \cdot (1+i)^{k-2}
                - \ldots - P}
        \hspace{1in} \\
        & = &   L (1+i)^k - P \cdot \frac{(1+i)^k - 1}{i}.
\end{eqnarray*}

The loan is paid when the amount that I owe equals zero; that is, when

\begin{eqnarray*}
L (1+i)^k - P \cdot \frac{(1+i)^k - 1}{i} & = & 0.
\end{eqnarray*}

Solving for $k$, the number of months until the loan is repaid, gives

\begin{eqnarray*}
k & = & \frac{\ln\left(\frac{P}{P-iL}\right)}{\ln(1 + i)}.
\end{eqnarray*}
}

\problempart Now let's consider an ``infrequent payment, late start''
repayment strategy.  Suppose that I pay $2P$ dollars to the bank every
two months, making the first payment two months after I take out the
loan.  (Thus, after 2 months I owe $L (1 + i)^2 - 2P$ dollars, and
after 4 months I owe $L (1+i)^4 - 2P \cdot (1 + i)^2 - 2P$ dollars.)
Give a closed-form expression for the number of months until the loan
in repaid.

\solution{
\textbf{Solution:}\\
After $k$ months, where $k$ is even, I owe

\begin{eqnarray*}
\lefteqn{L (1+i)^k - 2P \cdot (1+i)^{k-2} - 2P \cdot (1+i)^{k-4}
                - \ldots - 2P}
        \hspace{1in} \\
        & = &   L (1+i)^k - 2P \cdot \frac{(1+i)^k - 1}{(1+i)^2 - 1}.
\end{eqnarray*}

As before, the loan is repaid when the amount I owe equals zero.  This
happens when

\begin{eqnarray*}
k & = & \frac{\ln\left(\frac{2P}{2P-((1+i)^2-1)L}\right)}{\ln(1 + i)}.
\end{eqnarray*}
}

\problempart Finally, let's consider an ``infrequent payment, early
start'' strategy.  As before, I pay $2P$ dollars to the bank every two
months, but now I make the first payment one month after I take out
the loan.  (Thus, after 1 month I owe $L (1 + i) - 2P$ dollars, and
after 3 months I owe $L (1 + i)^3 - 2P \cdot (1+i)^2 - 2P$ dollars.)
Give a closed-form expression for the number of months until the loan
is repaid.

\solution{
\textbf{Solution:}\\
After $k$ months, where $k$ is odd, I owe

\begin{eqnarray*}
\lefteqn{L (1+i)^k - 2P \cdot (1+i)^{k-1} - 2P \cdot (1+i)^{k-3}
                - \ldots - 2P}
        \hspace{1in} \\
        & = &   L (1+i)^k - 2P \cdot \frac{(1+i)^{k+1} - 1}{(1+i)^2 - 1}.
\end{eqnarray*}

The loan is repaid when

\begin{eqnarray*}
k & = & \frac{\ln\left(\frac{2P}{2P(1+i)-((1+i)^2-1)L}\right)}{\ln(1 + i)}.
\end{eqnarray*}
}

\problempart Suppose that I take a loan of $L = \$100,000$.  The
monthly interest rate is $i = 0.005$.  I either pay the bank $P =
\$600$ every month or $2P = \$1,200$ every two months, depending on
the repayment strategy that I choose.  How long will repayment take
for each of the three strategies?

\solution{
\textbf{Solution:}\\
Substitution gives the following results:

\begin{tabular}{lcr}
frequent payment & \hspace{1in} & 360 months \\
infrequent payment, late start & & 362 months \\
infrequent payment, early start & & 357 months (k is odd)
\end{tabular}
}

\problempart In 50 words or fewer, explain the merits of frequent and
infrequent payment strategies.

\solution{
\textbf{Solution:}\\
Getting money to the bank early is more important than the
frequency of payments.}

\end{problemparts}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

                                         

\end{problems}

\end{document}                            

