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--- the_binomial_theorem [2020/11/03 20:19] – hkimscil
+++ the_binomial_theorem [2020/11/18 00:20] (current) – hkimscil
@@ Line 56: / Line 56: @@
 a^3 + 3a^2b &+& 3ab^2 + b^3
 \end{eqnarray*}
+\begin{eqnarray*}
+   &{\huge 1}&   \\
+ {\huge 1}a &+& {\huge 1}b \\
+{\huge 1}a^2 + {\huge 2}&ab& + {\huge 1}b^2 \\
+{\huge 1}a^3 + {\huge 3}a^2b &+& {\huge 3}ab^2 + {\huge 1}b^3
+\end{eqnarray*}
+{{  :pasted:20201103-202420.png  }}
+따라서 $ (a+b)^4 $는
+| a 지수는 4, 3, 2, 1, 0 순  |  $a^4$  |  $a^3$  |  $a^2$  |  $a^1$  |  1  |
+| b 지수는 0, 1, 2, 3, 4 순  |  $a^4$  |  $a^3b$  |  $a^2b^2$  |  $a^1b^3$  |  $b^4$  |
+| coefficient 는 1, 4, 6, 4, 1 순  |   $a^4$  |  $4a^3b$  |  $6a^2b^2$  |  $4a^1b^3$  |  $b^4$  |
+<WRAP info>
+그렇다면,
+$(a + b)^5 = {\text ?} $
+</WRAP>
+\begin{eqnarray*}
+&{0 \choose 0}&   \\
+&{1 \choose 0} \; {1 \choose 1}& \\
+&{2 \choose 0}\; {2 \choose 1}\; {2 \choose 2}& \\
+&{3 \choose 0}\; {3 \choose 1}\; {3 \choose 2}\; {3 \choose 3}& \\
+&{4 \choose 0}\; {4 \choose 1}\; {4 \choose 2}\; {4 \choose 3}\; {4 \choose 4}& \\
+&{5 \choose 0}\; {5 \choose 1}\; {5 \choose 2}\; {5 \choose 3}\; {5 \choose 4}\; {5 \choose 5} & \\
+\end{eqnarray*}
+\begin{eqnarray*}
+& {\large 1} &  \\
+& {\large 1\quad 1} & \\
+& {\large 1\quad 2\quad 1} & \\
+& {\large 1\quad 3\quad 3\quad 1} & \\
+& {\large 1\quad 4\quad 6\quad 4\quad 1} & \\
+& {\large 1\quad 5\quad 10\quad 10\quad 5\quad 1} & \\
+\end{eqnarray*}
+따라서
+\begin{eqnarray*}
+& a^5 + a^4 + a^3 + a^2 + a^1 + a^0 & \\
+& b^0 + b^1 + b^2 + b^3 + b^4 + b^5 & \\
+& 1   + 5   + 10  + 10  + 5   + 1 & \\
+& a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 & \\
+\end{eqnarray*}
+위를 종합해서 정리하면
+\begin{eqnarray*}
+\text{The binomial theorem} &  & \\
+(a + b)^{n} & = & \sum^{n}_{k=0}{{n}\choose{k}} a^{n-k} b^{k} \\
+\end{eqnarray*}
+예를 들면, 아래와 같다. ($n = 3$ 인경우)
+\begin{eqnarray*}
+(a + b)^{3} & = & \sum^{3}_{k=0}{{3}\choose{k}} a^{3-k} b^{k} \\
+& = & {{3}\choose{0}} a^{3-0} b^{0} + {{3}\choose{1}} a^{3-1} b^{1} + {{3}\choose{2}} a^{3-2} b^{2} + {{3}\choose{3}} a^{3-3} b^{3}  \\
+& = & 1 \cdot a^{3} b^{0} + 3 \cdot a^{2} b^{1} + 3 \cdot  a^{1} b^{2} + 1 \cdot  a^{0} b^{3}  \\
+& = & a^{3} + 3a^{2} b + 3ab^{2} + b^{3}
+\end{eqnarray*}