圆周之美

`\left(a+b\right)^{2}=a^{2}+2ab+b^{2}`

`\left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}`

`\left(a+b\right)^{4}=a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4}`

`\left(a+b\right)^{5}=a^{5}+5a^{4}b+10a^{3}b^{2}+10a^{2}b^{3}+5ab^{4}+b^{5}`

`\left(a+b\right)^{n}=\sum_{i=0}^{n}C_{n}^{i}a^{n-i}b^{i}`

`\left(a+b+c\right)^{n}=\sum_{i=0}^{n}C_{n}^{i}a^{n-i}b^{i}+\sum_{i=0}^{n}C_{n}^{i}a^{n-i}c^{i}+\sum_{i=0}^{n}C_{n}^{i}b^{n-i}c^{i}`

`\qquad \quad \quad \quad \quad \qquad \qquad \qquad \qquad \qquad 1`

`\qquad \quad \quad \quad \qquad \qquad \qquad \qquad \qquad 1\quad 1`

`\qquad \quad \quad \quad \qquad \qquad \qquad \qquad 1\qquad 2\qquad 1`

`\qquad \quad \quad \quad \qquad \qquad \qquad 1\qquad 3\qquad \quad 3\qquad 1`

`\qquad \quad \quad \quad \qquad \qquad 1\qquad 4\qquad \quad 6\qquad \quad 4\qquad 1`

`\qquad \quad \quad \qquad \qquad 1\qquad 5\qquad 10\qquad 10\qquad 5\qquad 1`

`\qquad \quad \quad \qquad 1\qquad 6\qquad 15\qquad 20\qquad 15\qquad 6\qquad 1`

`\qquad \quad \quad 1\qquad 7\qquad 21\qquad 35\qquad 35\qquad 21\qquad 7\qquad 1`

`\qquad \quad 1\qquad 8\qquad 28\qquad 56\qquad 70\qquad 56\qquad 28\qquad 8\qquad 1`

`\qquad 1\qquad 9\qquad 36\qquad 84\quad 126\quad 126\quad 84\qquad 36\qquad 9\qquad 1`

`1\qquad 10\quad 45\quad 120\quad 210\quad 252\quad 210\quad 120\quad 45\quad 10\qquad 1`

`a,b,c,d,e`

`ab \quad ac \quad ad \quad ae`

`bc \quad bd \quad be`

`cd \quad ce`

`de`

`共10种`

`C_5^2=10`

`a,b,c`

`ab \quad ac`

`bc`

`共3种`

`C_3^2=3`

`多项式定理：`

`{\sum_{i=0}^{n}C_{n}^{i}\sum_{j=1}^{m-1}B_{j}^{n-i}\sum_{k=j+1}^{m}B_{k}^{i}}_{（通项公式）}`

`其中，n表示多项式的次数`

`m表示多项式的项数`

`a被加了 \underline 4 次`

`b被加了 \underline 4 次`

`c被加了 \underline 4 次`

`d被加了 \underline 4 次`

`e被加了 \underline 4 次`

`a被加了 \underline 2 次`

`b被加了 \underline 2 次`

`c被加了 \underline 2 次`

`被多加的项：`

`\left(m-1\right)\sum_{l=1}^{m}B_l^n`

`最终公式：`

`\left(B_{1}+B_{2}+\cdot\cdot\cdot+B_{m}\right)^{n}=\sum_{i=0}^{n}C_{n}^{i}\sum_{j=1}^{m-1}B_{j}^{n-i}\sum_{k=j+1}^{m}B_{k}^{i}-\left(m-1\right)\sum_{l=1}^{m}B_{l}^{n}`

`令a=B_{1}, b=B_{2},c=B_{3},d=B_{4},e=B_{5}`

`\left(B_{1}+B_{2}+B_{3}+B_{4}+B_{5}\right)^{n}`

`=\sum_{i=0}^{n}C_{n}^{i}B_{1}^{n-i}B_{2}^{i}+\sum_{i=0}^{n}C_{n}^{i}B_{1}^{n-i}B_{3}^{i}+\sum_{i=0}^{n}C_{n}^{i}B_{1}^{n-i}B_{4}^{i}+\cdot\cdot\cdot`

`=\sum_{i=0}^{n}C_{n}^{i}\left(B_{1}^{n-i}B_{2}^{i}+B_{1}^{n-i}B_{3}^{i}+B_{1}^{n-i}B_{4}^{i}+\cdot\cdot\cdot\right)`

`=\sum_{i=0}^{n}C_{n}^{i}\left[\sum_{j=1}^{m-1}B_{j}^{n-i}\left(\sum_{k=j+1}^{m}B_{k}^{i}\right)\right]`

`好耶，`

`是“五项式定理”耶！`

`总项数：\left(n+1\right)\left(m-1\right)m`

`通项：C_{n}^{i}B_{p}^{n-i}B_{q}^{i}`

`令n=5`

`T_{1}=C_{5}^{0}B_{1}^{5}B_{2}^{0}`

`T_{2}=C_{5}^{0}B_{1}^{4}B_{2}^{1}`

`T_{3}=C_{5}^{0}B_{1}^{3}B_{2}^{2}`

`T_{4}=C_{5}^{0}B_{1}^{2}B_{2}^{3}`

`T_{5}=C_{5}^{0}B_{1}^{1}B_{2}^{4}`

`T_{6}=C_{5}^{0}B_{1}^{0}B_{2}^{5}`

`T_{7}=C_{5}^{0}B_{1}^{0}B_{3}^{5}`

`T_{8}=C_{5}^{0}B_{1}^{1}B_{3}^{4}`

`T_{9}=C_{5}^{0}B_{1}^{2}B_{3}^{3}`

`T_{10}=C_{5}^{0}B_{1}^{3}B_{3}^{2}`

`T_{11}=C_{5}^{0}B_{1}^{4}B_{3}^{1}`

`T_{12}=C_{5}^{0}B_{1}^{5}B_{3}^{0}`

`T_{13}=C_{5}^{0}B_{1}^{0}B_{4}^{5}`

`T_{14}=C_{5}^{0}B_{1}^{1}B_{4}^{4}`

`T_{15}=C_{5}^{0}B_{1}^{2}B_{4}^{3}`

`\cdot\cdot\cdot\cdot\cdot\cdot`

`余数是周期性变化的，所以应该取余。`

`\qquad \left(n+1\right)项\cdot\left(m-1\right)项\cdot m项`

`\qquad \qquad \qquad i \qquad \qquad \cdot \qquad \qquad j \qquad \qquad \cdot \qquad \qquad k \qquad \qquad `

`\qquad 1° \qquad 0 \qquad \qquad \cdot \qquad \qquad 1 \qquad \qquad \cdot \qquad \qquad 2 \qquad \qquad `

`\qquad 2° \qquad 0 \qquad \qquad \cdot \qquad \qquad 1 \qquad \qquad \cdot \qquad \qquad 3 \qquad \qquad `

`在m=5,n=5时`

`项数 6 \qquad \qquad \cdot \qquad \qquad 4 \qquad \qquad \cdot \qquad \qquad 5 \qquad \qquad _{=120}`

`i=向下取整\left(\frac{s}{\left(m-1\right)m}\right)`

`j=向上取整\left(\frac{s}{\left(n+1\right)m}\right)`

`k=向上取整\left(\frac{s}{\left(n+1\right)\left(m-1\right)}\right)`

`哦耶！通项公式来喽！`

`T_{s}=C_{n}^{向下取整\left(\frac{s}{\left(m-1\right)m}\right)} \cdot B_{向上取整\left(\frac{s}{\left(n+1\right)m}\right)}^{n-向下取整\left(\frac{s}{\left(m-1\right)m}\right)} \cdot B_{向上取整\left(\frac{s}{\left(n+1\right)\left(m-1\right)}\right)}^{n-向下取整\left(\frac{s}{\left(m-1\right)m}\right)}`

`（这个是错的）`

`s \quad 1,2,3,4,5,6,7,8,9,10,11,12`

`k \quad 1,2,3,4,5,6,1,2,3,4,5,6`

`Next\ Station:`

`余数与周期性.`

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