the_binomial_theorem
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| Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
| the_binomial_theorem [2020/11/04 16:10] – [계수] hkimscil | the_binomial_theorem [2020/11/18 00:20] (current) – hkimscil | ||
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| &{3 \choose 0}\; {3 \choose 1}\; {3 \choose 2}\; {3 \choose 3}& \\ | &{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}& \\ | &{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*} | \end{eqnarray*} | ||
| \begin{eqnarray*} | \begin{eqnarray*} | ||
| - | & {\large 1} & | + | & {\large 1} & \\ |
| - | & {\large 1\quad | + | & {\large 1\quad 1} & \\ |
| - | & {\large 1\quad | + | & {\large 1\quad 2\quad 1} & \\ |
| & {\large 1\quad 3\quad 3\quad 1} & \\ | & {\large 1\quad 3\quad 3\quad 1} & \\ | ||
| & {\large 1\quad 4\quad 6\quad 4\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*} | \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 | 위를 종합해서 정리하면 | ||
| Line 98: | Line 110: | ||
| \end{eqnarray*} | \end{eqnarray*} | ||
| + | 예를 들면, 아래와 같다. ($n = 3$ 인경우) | ||
| \begin{eqnarray*} | \begin{eqnarray*} | ||
| (a + b)^{3} & = & \sum^{3}_{k=0}{{3}\choose{k}} a^{3-k} b^{k} \\ | (a + b)^{3} & = & \sum^{3}_{k=0}{{3}\choose{k}} a^{3-k} b^{k} \\ | ||
the_binomial_theorem.1604473846.txt.gz · Last modified: by hkimscil