Follow the outline on the next page to use mathematical induction to prove
that
$$
\begin{aligned}(a+b)^{n}=\left(\begin{array}{c}n \\\0\end{array}\right)
a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1}
b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2}
\\\\+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a
b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned}
$$
a. Verify the formula for \(n=1\)
b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace
\(n\) with \(k+1\) and write the statement that must be proved.
c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add
exponents on the left. On the right, distribute \(a\) and \(b,\) respectively.
d. Collect like terms on the right. At this point, you should have
$$
\begin{aligned}&(a+b)^{k+1}=\left(\begin{array}{l}k \\\0\end{array}\right)
a^{k+1}+\left[\left(\begin{array}{l}k
\\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)\right]
a^{k} b\\\&\begin{array}{l}+\left[\left(\begin{array}{c}k
\\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right]
a^{k-1} b^{2}+\left[\left(\begin{array}{c}k
\\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right]
a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k
\\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right]
a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1}
\end{array}\end{aligned}
$$
e. Use the result of Exercise 74 to add the binomial sums in brackets. For
example, because \(\left(\begin{array}{l}n \\\
r\end{array}\right)+\left(\begin{array}{c}n \\\
r+1\end{array}\right)$$=\left(\begin{array}{l}n+1 \\ r+1\end{array}\right),\)
then \(\left(\begin{array}{l}k \\ 0\end{array}\right)+\left(\begin{array}{l}k
\\\ 1\end{array}\right)=\left(\begin{array}{c}k+1 \\\1\end{array}\right)\)
and\(\left(\begin{array}{l}k \\ 1\end{array}\right)+\left(\begin{array}{l}k
\\\2\end{array}\right)=\left(\begin{array}{c}k+1 \\ 2\end{array}\right)\)
f. Because \(\left(\begin{array}{l}k \\\
0\end{array}\right)=\left(\begin{array}{c}k+1 \\ 0\end{array}\right) \quad\)
(why?) and \(\left(\begin{array}{l}k \\ k\end{array}\right)=\)
\(\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these
results and the results from part (e) into the equation in part (d). This
should give the statement that we were required to prove in the second step of
the mathematical induction process.
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