Binomial Newton

 Pengayaan:

$\color{blue}\textrm{E. Binomial Newton}$

$\color{blue}\textrm{E. 1 Binomial Newton}$

$\begin{aligned}&\textrm{Perhatikanlah susunan bilangan berikut}\\\\ &\begin{array}{|c|l|}\hline &\\ 1=C_{0}^{\color{red}1}\quad 1=C_{1}^{\color{red}1}&(a+b)^{\color{red}1}\\ &\\ 1=C_{0}^{\color{red}2}\quad 2=C_{1}^{\color{red}2}\quad 1=C_{2}^{\color{red}2}&(a+b)^{\color{red}2}\\ &\\ 1=C_{0}^{\color{red}3}\quad 3=C_{1}^{\color{red}3}\quad 3=C_{2}^{\color{red}3}\quad 1=C_{3}^{\color{red}3}&(a+b)^{\color{red}3}\\ &\\ 1=C_{0}^{\color{red}4}\quad 4=C_{1}^{\color{red}4}\quad 6=C_{2}^{\color{red}4}\quad 4=C_{3}^{\color{red}4}\quad 1=C_{4}^{\color{red}4}&(a+b)^{\color{red}4}\\ \vdots &\: \: \quad\vdots \\ dst&(a+b)^{\color{red}\cdots }\\ &\\ \vdots&\: \: \quad\vdots \\ &(a+b)^{\color{red}n}\\\hline \end{array}\\\\ &\textrm{Susunan bilangan-bilangan di atas selanjutnya}\\ &\textrm{dinamakan}\: \: \: \textbf{Segitiga Pascal}\\ & \end{aligned}$

$\begin{aligned}&\textrm{Bilangan}\: \: C_{r}^{n}=\begin{pmatrix} n\\ r \end{pmatrix}\: \: \textrm{merupakan koefisien}\\ &\textrm{dari binomial}\: \: (a+b)^{n}\\ &\textrm{Selanjutnya perhatikanlah bahwa untuk}\\ &n=1,2,3,4,\cdots \: \: \: \textrm{berlaku}\\ &\color{red}\begin{aligned}(a+b)^{n}\color{black}=\, &\color{red}C_{0}^{n}a^{n}b^{0}+C_{1}^{n}a^{n-1}b^{1}+C_{2}^{n}a^{n-2}b^{2}\\ &+C_{3}^{n}a^{n-3}b^{3}+\cdots +C_{n-3}^{n}a^{3}b^{n-3}\\ &+C_{n-2}^{n}a^{2}b^{n-2}+C_{n-1}^{n}a^{1}b^{n-1}+C_{n}^{n}a^{0}b^{n}\\ &\color{black}=\displaystyle \sum_{r=0}^{n}C_{r}^{\color{red}n}a^{\color{red}n\color{black}-r}b^{r} \end{aligned}\\ & \end{aligned}$

$\color{blue}\textrm{E. 2 Perluasan Binomial Newton}$

$\begin{aligned}&\textrm{Untuk bilangan real}\: \: n\: \: \textrm{dan bilangan}\\ &\textrm{non negatif}\: \: r,\: \: \textrm{serta}\: \: \left | A \right |<1,\: \textrm{berlaku}:\\ &(1+A)^{n}=\displaystyle \sum_{r=0}^{n}C_{r}^{n}A^{r} \end{aligned}$

$\color{blue}\textrm{E. 3 Teorema Multinomial}$

Pada bentuk multinomial dengan ekspresi  $(x_{1}+x_{2}+x_{3}+\cdots +x_{r})^{n}$  dengan n dan r bilangan bulat positif, maka koefisien dari  $\color{red}x_{1}^{n_{1}}x_{2}^{n_{2}}x_{3}^{n_{3}}\cdots x_{r}^{n_{r}}$   adalah  $\displaystyle \frac{n!}{n_{1}!n_{2}!n_{3}!\cdots n_{r}!}$  dinotasikan dengan  $\begin{pmatrix} n\\\\ n_{1},n_{2},n_{3},\cdots ,n_{r} \end{pmatrix}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Misalkan untuk}\: \: n\: \: \textrm{bilangan bulat}\\ &\textrm{Positif. Tunjukklan bahwa}\\ &\textrm{a}.\quad (1+x)^{n}=\displaystyle \sum_{r=0}^{n}C_{r}^{n}x^{r}=\displaystyle \sum_{r=0}^{n}\begin{pmatrix} n\\ r \end{pmatrix}x^{r}\\ &\textrm{b}.\quad \begin{pmatrix} n\\ 0 \end{pmatrix}+\begin{pmatrix} n\\ 1 \end{pmatrix}+\begin{pmatrix} n\\ 2 \end{pmatrix}+\cdots +\begin{pmatrix} n\\ n \end{pmatrix}=2^{n}\\\\ &\textbf{Bukti}\\ &\color{red}\begin{aligned}\color{black}\textrm{a}.\quad(1+x)&^{n}\\ \color{black}=\, &\color{red}C_{0}^{n}1^{n}x^{0}+C_{1}^{n}1^{n-1}x^{1}+C_{2}^{n}1^{n-2}x^{2}\\ &+C_{3}^{n}1^{n-3}x^{3}+\cdots +C_{n-3}^{n}1^{3}x^{n-3}\\ &+C_{n-2}^{n}1^{2}x^{n-2}+C_{n-1}^{n}1^{1}x^{n-1}+C_{n}^{n}1^{0}x^{n}\\ =\, &\color{red}C_{0}^{n}+C_{1}^{n}x+C_{2}^{n}x^{2} +C_{3}^{n}x^{3}+\cdots \\ &+C_{n-3}^{n}x^{n-3} +C_{n-2}^{n}x^{n-2}+C_{n-1}^{n}x^{n-1}\\ &+C_{n}^{n}x^{n}\\ \color{black}\textrm{atau}&\: \color{black}\textrm{dengan bentuk lain}\\ =\, &\begin{pmatrix} n\\ 0 \end{pmatrix}+\begin{pmatrix} n\\ 1 \end{pmatrix}x+\begin{pmatrix} n\\ 2 \end{pmatrix}x^{2}+\begin{pmatrix} n\\ 3 \end{pmatrix}x^{3}\\ &+\cdots +\begin{pmatrix} n\\ n-3 \end{pmatrix}x^{n-3}+\begin{pmatrix} n\\ n-2 \end{pmatrix}x^{n-2}\\ &+\begin{pmatrix} n\\ n-1 \end{pmatrix}x^{n-1}+\begin{pmatrix} n\\ n \end{pmatrix}x^{n}\\ \color{black}=&\color{black}\displaystyle \sum_{r=0}^{n}\begin{pmatrix} \color{red}n\\ r \end{pmatrix}x^{r} \end{aligned}\\ &\color{red}\begin{aligned}\color{black}\textrm{b}.\quad(1+x)&^{n}\: \: \color{black}\textrm{lihat jawaban poin}\: \: a,\: \: \textrm{saat}\: \: \color{blue}x=1\\ \color{black}(1+1)&^{n}\color{red} \color{black}=\color{red}\begin{pmatrix} n\\ 0 \end{pmatrix}+\begin{pmatrix} n\\ 1 \end{pmatrix}1+\begin{pmatrix} n\\ 2 \end{pmatrix}1^{2}+\begin{pmatrix} n\\ 3 \end{pmatrix}1^{3}\\ &+\cdots +\begin{pmatrix} n\\ n-3 \end{pmatrix}1^{n-3}+\begin{pmatrix} n\\ n-2 \end{pmatrix}1^{n-2}\\ &+\begin{pmatrix} n\\ n-1 \end{pmatrix}1^{n-1}+\begin{pmatrix} n\\ n \end{pmatrix}1^{n}\\ \color{black}(2)&^{n}\color{red} \color{black}=\color{red}\begin{pmatrix} n\\ 0 \end{pmatrix}+\begin{pmatrix} n\\ 1 \end{pmatrix}+\begin{pmatrix} n\\ 2 \end{pmatrix}+\begin{pmatrix} n\\ 3 \end{pmatrix}\\ &+\cdots +\begin{pmatrix} n\\ n-1 \end{pmatrix}+\begin{pmatrix} n\\ n \end{pmatrix}\\ \color{black}=&\color{black}\displaystyle \sum_{r=0}^{n}\begin{pmatrix} n\\ r \end{pmatrix}\\ \color{black}\textrm{Sehing}&\color{black}\textrm{ga}\\ 2^{n}&=\displaystyle \sum_{r=0}^{n}\begin{pmatrix} n\\ r \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Misalkan untuk}\: \: n\: \: \textrm{bilangan bulat}\\ &\textrm{Positif. Tunjukklan bahwa}\\ & \begin{pmatrix} n\\ 0 \end{pmatrix}-\begin{pmatrix} n\\ 1 \end{pmatrix}+\begin{pmatrix} n\\ 2 \end{pmatrix}-\cdots +(-1)^{n}\begin{pmatrix} n\\ n \end{pmatrix}=0\\\\ &\textbf{Bukti}\\ &\textrm{Sebelumnya diketahui bahwa}\\ &\begin{aligned}&(a+b)^{n}=\displaystyle \sum_{r=0}^{n}\begin{pmatrix} n\\ r \end{pmatrix}a^{n-r}b^{r}\\ &\qquad\qquad\qquad \color{blue}\textrm{atau}\\ &\displaystyle \sum_{r=0}^{n}\begin{pmatrix} n\\ r \end{pmatrix}a^{n-r}b^{r}=(a+b)^{n}\\ &\blacklozenge \quad \textrm{saat}\: \: \color{blue}a=b=1,\: \: \color{black}\textrm{maka}\\ &\displaystyle \sum_{r=0}^{n}\begin{pmatrix} n\\ r \end{pmatrix}1^{n-r}1^{r}=(1+1)^{n}\\ &\Leftrightarrow \displaystyle \sum_{r=0}^{n}\begin{pmatrix} n\\ r \end{pmatrix}=2^{n}\: ...\: (\color{red}\textrm{bukti no. 1.b})\\ &\blacklozenge \quad \textrm{saat}\: \: \color{blue}a=1\: \&\: b=-1\: \: \color{black}\textrm{maka}\\ &\displaystyle \sum_{r=0}^{n}\begin{pmatrix} n\\ r \end{pmatrix}1^{n-r}(-1)^{r}=(1-1)^{n}=0\\ &\textrm{Sehingga}\\ &\begin{pmatrix} n\\ 0 \end{pmatrix}-\begin{pmatrix} n\\ 1 \end{pmatrix}+\begin{pmatrix} n\\ 2 \end{pmatrix}-\cdots +(-1)^{n}\begin{pmatrix} n\\ n \end{pmatrix}=0\quad \blacksquare \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Untuk}\: n,r\geq 0,\: \textrm{tunjukkan bahwa}\\ &\textrm{a}.\quad \begin{pmatrix} n\\ r \end{pmatrix}=\begin{pmatrix} n\\ n-r \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} n\\ r \end{pmatrix}=\displaystyle \frac{n}{r}\begin{pmatrix} n-1\\ r-1 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} n\\ r \end{pmatrix}=\displaystyle \frac{n-r+1}{r}\begin{pmatrix} n\\ r-1 \end{pmatrix}\\ &\textrm{d}.\quad\begin{pmatrix} -n\\ r \end{pmatrix}=(-1)^{k}\begin{pmatrix} n+r-1\\ r \end{pmatrix}\\ &\textrm{e}.\quad\begin{pmatrix} n\\ r \end{pmatrix}+\begin{pmatrix} n\\ r+1 \end{pmatrix}=\begin{pmatrix} n+1\\ r+1 \end{pmatrix}\\ &\textrm{f}.\quad \begin{pmatrix} n\\ m \end{pmatrix}\begin{pmatrix} m\\ r \end{pmatrix}=\begin{pmatrix} n\\ r \end{pmatrix}\begin{pmatrix} n-r\\ m-r \end{pmatrix}\\\\ &\textbf{Bukti}:\\ &\begin{aligned}\textrm{a}.\quad\begin{pmatrix} n\\ r \end{pmatrix}&=\displaystyle \frac{n!}{r!(n-r)!}\\ &=\displaystyle \frac{n!}{(n-r)!(n-(n-r))!}\\ &=\frac{n!}{(n-r)!r!}=\begin{pmatrix} n\\ n-r \end{pmatrix} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad\begin{pmatrix} n\\ r \end{pmatrix}&=\displaystyle \frac{n!}{r!(n-r)!}\\ &=\displaystyle \frac{n.(n-1)!}{r.(r-1)!\left ((n-1)-(r-1) \right )!}\\ &=\displaystyle \frac{n}{r}\frac{(n-1)!}{(r-1)!\left ( (n-1)-(r-1) \right )!}\\ &=\displaystyle \frac{n}{r}\begin{pmatrix} n-1\\ r-1 \end{pmatrix} \end{aligned}\\ &\begin{aligned}\textrm{c}.\quad\begin{pmatrix} n\\ r \end{pmatrix}&=\displaystyle \frac{n!}{r!(n-r)!}\\ &=\displaystyle \frac{n!}{r.(r-1)!(n-r)!}\times \frac{((n-r)+1)}{((n-r)+1)}\\ &=\displaystyle \frac{n-r+1}{r}\times \frac{n!}{(r-1)!((n-r)+1)!}\\ &=\displaystyle \frac{n-r+1}{r}\times \frac{n!}{(r-1)!(n-(r-1))!}\\ &=\displaystyle \frac{n-r+1}{r}\begin{pmatrix} n\\ r-1 \end{pmatrix} \end{aligned}\\ &\textrm{d}.\quad\textrm{Silahkan dicoba buat latihan}\\ &\textrm{e}.\quad\textrm{Silahkan dicoba buat latihan}\\ &\textrm{f}.\quad\textrm{Silahkan dicoba buat latihan} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Tentukan nilai dari}\\ &\textrm{a}.\quad \begin{pmatrix} 1\\ 1 \end{pmatrix}+\begin{pmatrix} 2\\ 1 \end{pmatrix}+\begin{pmatrix} 3\\ 1 \end{pmatrix}+\cdots +\begin{pmatrix} 100\\ 1 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 100\\ 100 \end{pmatrix}+\begin{pmatrix} 101\\ 100 \end{pmatrix}+\begin{pmatrix} 102\\ 100 \end{pmatrix}+\cdots +\begin{pmatrix} 200\\ 100 \end{pmatrix}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\textrm{a}.\quad &\begin{pmatrix} 1\\ 1 \end{pmatrix}+\begin{pmatrix} 2\\ 1 \end{pmatrix}+\begin{pmatrix} 3\\ 1 \end{pmatrix}+\cdots +\begin{pmatrix} 100\\ 1 \end{pmatrix}\\ &\color{red}\textrm{Sebelumnya perhatikan}\\ &=\begin{pmatrix} 1\\ 1 \end{pmatrix}+\begin{pmatrix} 2\\ 1 \end{pmatrix}+\begin{pmatrix} 3\\ 1 \end{pmatrix}+\cdots +\begin{pmatrix} n\\ 1 \end{pmatrix}\\ &\textrm{Karena}\\ &\begin{pmatrix} n\\ r-1 \end{pmatrix}+\begin{pmatrix} n\\ r \end{pmatrix}=\begin{pmatrix} n+1\\ r \end{pmatrix}\\ &\textrm{Saat}\\ &\begin{pmatrix} 1\\ 1 \end{pmatrix}+\begin{pmatrix} 2\\ 1 \end{pmatrix}=\begin{pmatrix} 2\\ 2 \end{pmatrix}+\begin{pmatrix} 2\\ 1 \end{pmatrix}=\begin{pmatrix} 3\\ 2 \end{pmatrix}\\ &\textrm{Sehingga}\\ &\underset{\begin{pmatrix} 5\\ 2 \end{pmatrix}}{\underbrace{\underset{\begin{pmatrix} 4\\ 2 \end{pmatrix}}{\underbrace{\underset{\begin{pmatrix} 3\\ 2 \end{pmatrix}}{\underbrace{\begin{pmatrix} 1\\ 1 \end{pmatrix}+\begin{pmatrix} 2\\ 1 \end{pmatrix}}}+\begin{pmatrix} 3\\ 1 \end{pmatrix} }}+\begin{pmatrix} 4\\ 1 \end{pmatrix}}}\\ &\textrm{maka}\\ &=\begin{pmatrix} 1\\ 1 \end{pmatrix}+\begin{pmatrix} 2\\ 1 \end{pmatrix}+\begin{pmatrix} 3\\ 1 \end{pmatrix}+\cdots +\begin{pmatrix} n\\ 1 \end{pmatrix}=\begin{pmatrix} n+1\\ 2 \end{pmatrix}\\ &\textrm{Jadi},\\ &\color{black}\begin{pmatrix} 1\\ 1 \end{pmatrix}+\begin{pmatrix} 2\\ 1 \end{pmatrix}+\begin{pmatrix} 3\\ 1 \end{pmatrix}+\cdots +\begin{pmatrix} 100\\ 1 \end{pmatrix}=\begin{pmatrix} 101\\ 2 \end{pmatrix} \end{aligned}\\ &\textrm{b}\quad \textrm{Silahkan coba sendiri}\\ &\: \: \quad\textrm{sebagai latihan} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad \displaystyle \sum_{r=0}^{1000}\begin{pmatrix} 1000\\ r \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 2009\\ 1 \end{pmatrix}+\begin{pmatrix} 2009\\ 2 \end{pmatrix}+\begin{pmatrix} 2009\\ 3 \end{pmatrix}+\cdots +\begin{pmatrix} 2009\\ 2004 \end{pmatrix}\\\\ &\: \: \qquad (\textbf{OSK 2009})\\\\ &\textrm{Jawab}:\\ &\textrm{a}.\quad \displaystyle \sum_{r=0}^{1000}\begin{pmatrix} 1000\\ r \end{pmatrix}=\displaystyle \sum_{r=0}^{1000}\begin{pmatrix} 1000\\ r \end{pmatrix}1^{1000-r}1^{r}\\\\ &\: \: \qquad=(1+1)^{1000}=2^{1000}\\ &\textrm{b}.\quad \displaystyle \sum_{r=0}^{2009}\begin{pmatrix} 2009\\ r \end{pmatrix}=2^{2009}\\ &\: \: \qquad \textrm{karena}\: \: \color{blue}\begin{pmatrix} n\\ r \end{pmatrix}\color{black}=\color{red}\begin{pmatrix} n\\ n-r \end{pmatrix},\: \color{black}\textrm{maka}\\ &\: \: \qquad \begin{pmatrix} 2009\\ 0 \end{pmatrix}=\begin{pmatrix} 2009\\ 2009 \end{pmatrix},\: \begin{pmatrix} 2009\\ 1 \end{pmatrix}=\begin{pmatrix} 2009\\ 2008 \end{pmatrix},\\ &\: \: \qquad \cdots ,\: \begin{pmatrix} 2009\\ 1004 \end{pmatrix}=\begin{pmatrix} 2009\\ 1005 \end{pmatrix}\\ &\: \: \qquad \textrm{Sehingga}\\ &\: \: \qquad \begin{pmatrix} 2009\\ 0 \end{pmatrix}+\begin{pmatrix} 2009\\ 1 \end{pmatrix}+\begin{pmatrix} 2009\\ 2 \end{pmatrix}+\cdots +\begin{pmatrix} 2009\\ 2009 \end{pmatrix}=\color{red}2^{2009}\\ &\: \: \qquad \Leftrightarrow \: \begin{pmatrix} 2009\\ 0 \end{pmatrix}+\begin{pmatrix} 2009\\ 1 \end{pmatrix}+\begin{pmatrix} 2009\\ 2 \end{pmatrix}+\cdots +\begin{pmatrix} 2009\\ 1004 \end{pmatrix}=\displaystyle \frac{2^{2009}}{2}=2^{2008}\\ &\: \: \qquad \Leftrightarrow \: 1+\begin{pmatrix} 2009\\ 1 \end{pmatrix}+\begin{pmatrix} 2009\\ 2 \end{pmatrix}+\begin{pmatrix} 2009\\ 3 \end{pmatrix}+\cdots +\begin{pmatrix} 2009\\ 1004 \end{pmatrix}=2^{2008}\\ &\: \: \qquad \Leftrightarrow \: \begin{pmatrix} 2009\\ 1 \end{pmatrix}+\begin{pmatrix} 2009\\ 2 \end{pmatrix}+\begin{pmatrix} 2009\\ 3 \end{pmatrix}+\cdots +\begin{pmatrix} 2009\\ 1004 \end{pmatrix}=2^{2008}-1 \end{array}$

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