\documentstyle[11pt]{article}
\begin{document}

\noindent
September 9, 1997

\bigskip
I wish to make a remark concerning the shifting of indices in a
sum. A sum:
$$\sum_{k=0}^{n-1} c_{k}$$
can also be written as:
$$\sum_{k=1}^{n} c_{k-1}.$$
This simple trick is sometimes useful as we shall see in the
proof of the binomial theorem.

\bigskip
{\bf Solutions} (\# 4 p.44)
Use induction to prove the binomial theorem:
$$(a + b)^{n} = \sum_{k=0}^{n} \left( \begin{array}{c} n \\ 
k \end{array} \right) a^{k}b^{n-k}.$$
\noindent
{\it Proof.} If $n = 1, LHS = a + b,$
$$ RHS = \left( 
\begin{array}{cc} 1 \\ 
0 \end{array} \right) a^{0}b^{1} + \left( \begin{array}{c} 1 \\ 
1 \end{array} \right) a^{1}b^{0} = a + b = LHS.$$
Assume that the Theorem is valid for n we want to show that it
is also valid for $n + 1$:
$$(a + b)^{n+1} = \sum_{k=0}^{n+1} \left( \begin{array}{c} n+1
\\  k \end{array} \right) a^{k}b^{n+1-k}.$$
We have:
\begin{eqnarray*}
LHS &=& (a + b)(a + b)^{n}\\
&=& (a + b)
\sum_{k=0}^{n}
\left(
\begin{array}{c} n \\  k \end{array} \right) a^{k}b^{n-k}\\
&=& \sum_{k=0}^{n} \left( \begin{array}{c} n \\ 
k \end{array} \right) a^{k+1}b^{n-k} + \sum_{k=0}^{n} \left(
\begin{array}{c} n \\  k \end{array} \right) a^{k}b^{n+1-k}.
\end{eqnarray*}
The first sum on the right above can be written as:
$$\sum_{k=0}^{n} \left( \begin{array}{c} n \\ 
k \end{array} \right) a^{k+1}b^{n-k} = a^{n+1} + \sum_{k=0}^{n-1}
\left( \begin{array}{cc} n \\ 
k \end{array} \right) a^{k+1}b^{n-k}.$$
The second sum on the right above can be written as:
$$\sum_{k=0}^{n} \left(
\begin{array}{c} n \\  k \end{array} \right) a^{k}b^{n+1-k}
= \sum_{k=1}^{n} \left(
\begin{array}{c} n \\  k \end{array} \right) a^{k}b^{n+1-k}
+ b^{n+1}.$$
Thus
$$LHS = a^{n+1} + \sum_{k=0}^{n-1}
\left( \begin{array}{cc} n \\ 
k \end{array} \right) a^{k+1}b^{n-k} + \sum_{k=1}^{n} \left(
\begin{array}{c} n \\  k \end{array} \right) a^{k}b^{n+1-k}
+ b^{n+1}.$$
We use the trick of shifting indices to write the first sum
above as:
\begin{eqnarray*}
\sum_{k=0}^{n-1}
\left( \begin{array}{cc} n \\ 
k \end{array} \right) a^{k+1}b^{n-k} &=& \sum_{k=1}^{n}
\left( \begin{array}{c} n \\ 
k-1 \end{array} \right) a^{k}b^{n-(k-1)}\\
&=& \sum_{k=1}^{n}
\left( \begin{array}{c} n \\ 
k-1 \end{array} \right) a^{k}b^{n+1-k}.
\end{eqnarray*}
Thus
\begin{eqnarray*}
(a + b)^{n} &=& a^{n+1} + \sum_{k=1}^{n}
\left( \begin{array}{c} n \\ 
k-1 \end{array} \right) a^{k}b^{n+1-k} + \sum_{k=1}^{n} \left(
\begin{array}{c} n \\  k \end{array} \right) a^{k}b^{n+1-k}
+ b^{n+1}\\
&=& a^{n+1} + \sum_{k=1}^{n}
\{\left( \begin{array}{c} n \\ 
k-1 \end{array} \right) + \left(
\begin{array}{c} n \\  k \end{array} \right)\} a^{k}b^{n+1-k}
+ b^{n+1}\\
&=& a^{n+1} + \sum_{k=1}^{n}
\left( \begin{array}{c} n+1 \\ 
k \end{array} \right) a^{k}b^{n+1-k}
+ b^{n+1}\\
&=& \sum_{k=0}^{n+1}
\left( \begin{array}{c} n+1 \\ 
k \end{array} \right) a^{k}b^{n+1-k}.
\end{eqnarray*}
We have used the identity:
$$\left( \begin{array}{c} n+1 \\ 
k \end{array} \right) = \left(
\begin{array}{c} n \\  k-1 \end{array} \right) + \left(
\begin{array}{c} n \\  k \end{array} \right)$$
which is verified by a direct calculation:
\begin{eqnarray*}
\left(
\begin{array}{c} n \\  k-1 \end{array} \right) + \left(
\begin{array}{c} n \\  k \end{array} \right) &=& {n! \over
(k-1)!(n-k+1)!} + {n! \over
k!(n-k)!} \\
&=& {n!k + n!(n-k+1) \over
k!(n-k+1)!}\\
&=& {(n+1)! \over
k!(n-k+1)!}\\
&=& \left( \begin{array}{c} n+1 \\ 
k \end{array} \right).
\end{eqnarray*}
If we set $a = b = 1$ in the binomial theorem we obtain the
identity:
$$2^{n} = \sum_{k=0}^{n} \left( \begin{array}{c} n \\ 
k \end{array} \right).$$
On the other hand, if we take $a = -1$ and $b = 1$ then we get
$$\sum_{k=0}^{n} (-1)^{k} \left( \begin{array}{c} n \\ 
k \end{array} \right) = 0.$$

\bigskip
\noindent
HW p.43 \# 3, p.44 \# 5, 6, 7 p.56 \# 3, 9

\end{document}