the_binomial_theorem
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the_binomial_theorem [2020/11/03 20:30] – hkimscil | the_binomial_theorem [2020/11/18 00:20] (current) – hkimscil | ||
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따라서 $ (a+b)^4 $는 | 따라서 $ (a+b)^4 $는 | ||
| a 지수는 4, 3, 2, 1, 0 순 | $a^4$ | $a^3$ | $a^2$ | $a^1$ | 1 | | | 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 | $a^4$ | $a^3b$ | + | | b 지수는 0, 1, 2, 3, 4 순 | $a^4$ | $a^3b$ |
- | | coefficient 는 1, 4, 6, 4, 1 순 | | + | | coefficient 는 1, 4, 6, 4, 1 순 | |
+ | |||
+ | <WRAP info> | ||
+ | 그렇다면, | ||
+ | $(a + b)^5 = {\text ?} $ | ||
+ | </ | ||
+ | |||
+ | \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*} | ||
the_binomial_theorem.1604403056.txt.gz · Last modified: 2020/11/03 20:30 by hkimscil