08. 이항정리

이항정리

파스칼의 삼각형

$\begin{array}{c}(a+b{)}^0\\ (a+b{)}^1\\ (a+b{)}^2\\ (a+3{)}^3\\ (a+b{)}^4\\ (a+b{)}^5\\ ⋮\end{array}\begin{array}{c}1\\ 11\\ 121\\ 1331\\ 14641\\ 15101051\\ ⋮\end{array}$$\begin{matrix} (a+b)^0 \\\ (a+b)^1 \\\ (a+b)^2 \\\ (a+3)^3 \\\ (a+b)^4 \\\ (a+b)^5 \\\ \vdots\end{matrix} \qquad\begin{gather}1 \\\ \color{red}1 \qquad \color{red}1\\\ \color{blue}1\qquad \color{blue}2\qquad \color{blue}1 \\\ \color{silver}1 \qquad \color{silver}3\qquad \color{silver}3 \qquad \color{silver}1 \\\ \color{maroon}1\qquad \color{maroon}4\qquad \color{maroon}6\qquad \color{maroon}4\qquad \color{maroon}1 \\\ \color{lime}1\qquad \color{lime}5\qquad \color{lime}{10} \qquad \color{lime}{10} \qquad \color{lime}5 \qquad \color{lime}1\\\ \vdots\end{gather}$

$\Downarrow$$\qquad\qquad\qquad\qquad \Downarrow$

$\begin{array}{c}\begin{array}{c}{}_0{C}_0\\ {}_1{C}_0{}_1{C}_1\\ {}_2{C}_0{}_2{C}_1{}_2{C}_2\\ {}_3{C}_0{}_3{C}_1{}_3{C}_2{}_3{C}_3\\ {}_4{C}_0{}_4{C}_1{}_4{C}_2{}_4{C}_3{}_4{C}_4\\ {}_5{C}_0{}_5{C}_1{}_5{C}_2{}_5{C}_3{}_5{C}_4{}_5{C}_5\\ ⋮\end{array}\end{array}$$\qquad\quad\begin{gather} _0C_0 \\\ _1C_0\qquad _1C_1 \\\ _2C_0\qquad _2C_1 \qquad _2C_2 \\\ _3C_0 \qquad _3C_1 \qquad _3C_2 \qquad _3C_3 \\\ _4C_0 \qquad _4C_1 \qquad _4C_2 \qquad _4C_3 \qquad _4C_4 \\\ _5C_0 \qquad _5C_1 \qquad _5C_2 \qquad _5C_3 \qquad _5C_4 \qquad _5C_5\\\ \vdots \end{gather}$

$\begin{array}{rl}(a+b{)}^1& =1a+1b\\ (a+b{)}^2& =1a^2+2ab+1b^2\\ (a+b{)}^3& =1a^3+3a^{2}b+3a{b}^2+1b^3\\ (a+b{)}^4& =1a^4+4a^{3}b+6a^2{b}^{2}4a^{3}b+1b^4\\ (a+b{)}^5& =1a^5+5a^{4}b+10a^3{b}^2+10a^2{b}^3+5a{b}^4+1b^5\\ & ⋮\end{array}$$\begin{align} (a+b)^1&=\color{red}1a+\color{red}1b \\\ (a+b)^2 &=\color{blue}1a^2 +\color{blue}2ab+\color{blue}1b^2 \\\ (a+b)^3 &=\color{silver}1 a^3 +\color{silver}3 a^2 b +\color{silver}3ab^2 +\color{silver}1b^3 \\\ (a+b)^4 &=\color{maroon}1a^4 +\color{maroon}4 a^3b+\color{maroon}6 a^2b^2 \color{maroon}4a^3 b +\color{maroon}1b^4 \\\ (a+b)^5&= \color{lime}1 a^5+\color{lime}5 a^4b+\color{lime}{10}a^3b^2 +\color{lime}{10}a^2b^3 +\color{lime}5ab^4 +\color{lime}1b^5\\\ &\qquad\qquad \vdots\end{align}$

이를 좀 더 일반화 하면,

$\begin{array}{rl}(a+b{)}^n& =\sum _{r=0}^n{}_n{C}_r(a{)}^{n-r}(b{)}^r\\ & {=}_n{C}_0{a}^n{b}^0{+}_n{C}_1{a}^{n-1}{b}^1+\cdots {+}_n{C}_{n-1}{a}^1{b}^{n-1}{+}_n{C}_n{a}^0{b}^n\end{array}$$\begin{align} (a+b)^n&=\sum\limits_{r=0}^n {}_nC_r (a)^{n-r}(b)^r\\\ &=_nC_0 a^nb^0 +_nC_1 a^{n-1}b^1+\cdots+_nC_{n-1}a^1b^{n-1}+_nC_n a^0b^n\end{align}$

이 됩니다. 반드시 기억해야한다!

이항정리의 활용

$(1+x{)}^n=\sum _{r=1}^n{}_n{C}_r{x}^r$$(1+x)^n =\sum\limits_{r=1}^n {}_nC_rx^r$

을 이용하면,

원래의 식에

I. $x=1$$~x=1$을 대입하면,

$\begin{array}{rl}{2}^n& =\sum _{r=0}^n{}_n{C}_r\\ & {=}_n{C}_0{+}_n{C}_1{+}_n{C}_2+\cdots {+}_n{C}_n\end{array}$$\begin{align}2^n &=\sum\limits_{r=0}^n {}_nC_r\\\ &= _nC_0+_nC_1+_nC_2+\cdots +_nC_n\end{align}$

II. $x=-1$$~x=-1$을 대입하면,

$\begin{array}{rl}0& =\sum _{r=0}^n{}_n{C}_r(-1{)}^r\\ & {=}_n{C}_0{-}_n{C}_1{+}_n{C}_2+\cdots {+}_n{C}_r(-1{)}^r\end{array}$$\begin{align} 0&=\sum\limits_{r=0}^n {}_nC_r (-1)^r \\\ &=_nC_0 -_nC_1 +_nC_2+\cdots+_nC_r(-1)^r\end{align}$

$\Rightarrow {}_n{C}_0{+}_n{C}_2{+}_n{C}_4+\cdots {=}_n{C}_1{+}_n{C}_3{+}_n{C}_5+\cdots$$\Rightarrow~~_nC_0+_nC_2+_nC_4 +\cdots =_nC_1 +_nC_3+_nC_5+\cdots$

III. I과 II를 이용하면,

$\Rightarrow \underset{A}{\underbrace{{}_n{C}_0{+}_n{C}_2{+}_n{C}_4+\cdots }}=\underset{B}{\underbrace{{}_n{C}_1{+}_n{C}_3{+}_n{C}_5+\cdots }}$$\Rightarrow~~\underset{A}{\underbrace{_nC_0+_nC_2+_nC_4 +\cdots }}=\underset{B}{\underbrace{_nC_1 +_nC_3+_nC_5+\cdots}}$

$A=B$$A=B$ 그리고, $A+B=2^n$$~A+B=2^n$

$\therefore A=B=2^{n-1}$$\therefore~~A=B=2^{n-1}$

미분한 후에

$n(1+x{)}^{n-1}=\sum _{r=0}^{n}r\cdot {}_n{C}_r{x}^{r-1}$$n(1+x)^{n-1} =\sum\limits_{r=0}^n r \cdot{}_nC_r x^{r-1}$

I. $x=1$$~x=1$을 대입하면,

$\begin{array}{rl}n\cdot 2^{n-1}& =\sum _{r=0}^{n}r{\cdot }_n{C}_r\\ & =0{\cdot }_n{C}_0+1{\cdot }_{n}C1+2{\cdot }_n{C}_2+\cdots +n{\cdot }_n{C}_n\end{array}$$\begin{align}n\cdot2^{n-1}&=\sum\limits_{r=0}^n r\cdot _nC_r \\\ &= 0\cdot _nC_0 +1\cdot _nC1 +2\cdot _nC_2+\cdots+n\cdot _nC_n \end{align}$

II. $x=-1$$~x=-1$을 대입하면,

$\begin{array}{rl}0& =\sum _{r=0}^{n}r\cdot {}_n{C}_r(-1{)}^r\\ & =0\cdot {}_n{C}_0(-1{)}^0-1\cdot {}_n{C}_1+2\cdot {}_n{C}_2-3\cdot {}_n{C}_3+4\cdot {}_n{C}_4+\cdots \\ & ⋮\end{array}$$\begin{aligned} 0&=\sum\limits_{r=0}^n r\cdot {}_nC_r (-1)^r \\\ &=0\cdot {}_nC_0(-1)^0 -1\cdot {}_nC_1 +2\cdot {}_nC_2 -3\cdot {}_nC_3 +4\cdot {}_nC_4 +\cdots\\ &\quad\quad\quad\vdots\end{aligned}$

$1{\cdot }_n{C}_1+3{\cdot }_n{C}_3+5{\cdot }_n{C}_5+\cdots =2{\cdot }_n{C}_2+4{\cdot }_n{C}_4+6{\cdot }_n{C}_6+\cdots$$1\cdot _nC_1 +3\cdot _nC_3 +5\cdot _nC_5 +\cdots =2\cdot _nC_2 +4\cdot _nC_4+6\cdot _nC_6+\cdots$

III. I과 II를 이용하면,

$\underset{A}{\underbrace{1{\cdot }_n{C}_1+3{\cdot }_n{C}_3+5{\cdot }_n{C}_5+\cdots }}=\underset{B}{\underbrace{2{\cdot }_n{C}_2+4{\cdot }_n{C}_4+6{\cdot }_n{C}_6+\cdots }}$$\underset{A}{\underbrace{1\cdot _nC_1 +3\cdot _nC_3 +5\cdot _nC_5 +\cdots}} =\underset{B}{\underbrace{2\cdot _nC_2 +4\cdot _nC_4+6\cdot _nC_6+\cdots}}$

$A=B$$A=B$ 그리고, $A+B=n\cdot 2^{n-1}$$~A+B=n\cdot 2^{n-1}$

$\therefore A=B=\frac{n\cdot 2^{n-1}}{2}=n\cdot 2^{n-2}$$\therefore~~A=B=\cfrac {n\cdot 2^{n-1}}2 =n\cdot 2^{n-2}$

적분한 후에

$\frac{1}{n+1}(1+x{)}^{n+1}=\sum _{r=0}^n\frac{1}{r+1}{}_n{C}_r{x}^{r+1}$$\cfrac 1 {n+1} (1+x)^{n+1}=\sum\limits_{r=0}^n \cfrac 1 {r+1} {}_nC_r x^{r+1}$

I. $x=1$$~x=1$을 대입하면,

$\frac{1}{n+1}{2}^{n+1}=\sum _{r=0}^n\frac{1}{r+1}{}_n{C}_r$$\cfrac 1 {n+1} 2^{n+1} =\sum\limits_{r=0}^n \cfrac 1 {r+1} {}_nC_r$

$\frac{1}{n+1}{2}^{n+1}=\frac{1}{1}{}_n{C}_0+\frac{1}{2}{}_n{C}_1+\frac{1}{3}{}_n{C}_2+\cdots +\frac{1}{n+1}{}_n{C}_n$$\cfrac 1 {n+1} 2^{n+1} = \cfrac 1 1 {}_nC_0 + \cfrac 1 2 {}_nC_1+\cfrac 1 3 {}_nC_2+\cdots+\cfrac 1{n+1} {}_nC_n$

II. $x=-1$$~x=-1$을 대입하면,

$0=-\frac{1}{1}{}_n{C}_0+\frac{1}{2}{}_n{C}_1-\frac{1}{3}{}_n{C}_2+\cdots +\frac{1}{n+1}{}_{n}Cn(-1{)}^{n+1}$$0= -\cfrac 1 1 {}_nC_0 +\cfrac 1 2 {}_nC_1-\cfrac 1 3 {}_nC_2 +\cdots +\cfrac 1 {n+1} {}_nCn(-1)^{n+1}$

${}_n{C}_0+\frac{1}{3}{}_n{C}_2+\frac{1}{5}{}_n{C}_4+\cdots =\frac{1}{2}{}_n{C}_1+\frac{1}{4}{}_n{C}_3+\frac{1}{6}{}_n{C}_5+\cdots$$_nC_0 +\cfrac 1 3{}_nC_2 +\cfrac 1 5 {}_nC_4+\cdots =\cfrac 1 2 {}_nC_1 +\cfrac 1 4 {}_nC_3 +\cfrac 1 6 {}_nC_5+\cdots$

III. I과 II를 이용하면,

$\underset{A}{\underbrace{{}_n{C}_0+\frac{1}{3}{}_n{C}_2+\frac{1}{5}{}_n{C}_4+\cdots }}=\underset{B}{\underbrace{\frac{1}{2}{}_n{C}_1+\frac{1}{4}{}_n{C}_3+\frac{1}{6}{}_n{C}_5+\cdots }}$$\underset{A}{\underbrace{_nC_0 +\cfrac 1 3{}_nC_2 +\cfrac 1 5 {}_nC_4+\cdots }}=\underset{B}{\underbrace{\cfrac 1 2 {}_nC_1 +\cfrac 1 4 {}_nC_3 +\cfrac 1 6 {}_nC_5+\cdots}}$

$A=B$$A=B$ 그리고, $A+B=\frac{1}{n+1}{2}^{n+1}$$~A+B= \cfrac 1 {n+1} 2^{n+1}$

$\therefore A=B=\frac{1}{n+1}{2}^{n+1}\times \frac{1}{2}=\frac{1}{n+1}{2}^n$$\therefore ~~ A=B= \cfrac 1 {n+1} 2^{n+1} \times \cfrac 1 2 =\cfrac 1 {n+1} 2^n$

 

정리

• $(a+b{)}^n=\sum _{r=0}^n{}_n{C}_r{a}^{n-r}{b}^r$$~(a+b)^n =\sum\limits_{r=0}^n {}_nC_r a^{n-r}b^r$

• $(1+x{)}^n=\sum _{r=0}^n{}_n{C}_r{x}^r$$(1+x)^n =\sum\limits_{r=0}^n {}_nC_r x^r$

$\begin{array}{c}⟹\{\begin{array}{l}\bullet \text{그대로}\{\begin{array}{l}x=1\\ x=-1\end{array}\\ \\ \bullet \text{미분한 후}\{\begin{array}{l}x=1\\ x=-1\end{array}\\ \\ \bullet \text{적분한 후}\{\begin{array}{l}x=1\\ x=-1\end{array}\end{array}\end{array}$$\Longrightarrow ~~\begin{cases} \bullet~~\text{그대로}~~\begin{cases} x=1\\\ x=-1\end{cases} \\\ \\\ \bullet~~\text{미분한 후}~~\begin{cases} x=1 \\\ x=-1\end{cases} \\\ \\\ \bullet~~\text{적분한 후}~~\begin{cases} x=1 \\\ x=-1 \end{cases}\end{cases}$