PAS MATEMATIKA BLOG: Binomial Newton

Pengayaan:

$E. Binomial Newton$

$E. 1 Binomial Newton$

$\begin{aligned} Perhatikanlah susunan bilangan berikut \\ \begin{array}{cl} 1 = C_{0}^{1} 1 = C_{1}^{1} & (a + b)^{1} \\ 1 = C_{0}^{2} 2 = C_{1}^{2} 1 = C_{2}^{2} & (a + b)^{2} \\ 1 = C_{0}^{3} 3 = C_{1}^{3} 3 = C_{2}^{3} 1 = C_{3}^{3} & (a + b)^{3} \\ 1 = C_{0}^{4} 4 = C_{1}^{4} 6 = C_{2}^{4} 4 = C_{3}^{4} 1 = C_{4}^{4} & (a + b)^{4} \\ ⋮ & ⋮ \\ d s t & (a + b)^{\dots} \\ ⋮ & ⋮ \\ (a + b)^{n} \end{array} \\ Susunan bilangan-bilangan di atas selanjutnya \\ dinamakan Segitiga Pascal \end{aligned}$

$\begin{aligned} Bilangan C_{r}^{n} = (\begin{array}{c} n \\ r \end{array}) merupakan koefisien \\ dari binomial (a + b)^{n} \\ Selanjutnya perhatikanlah bahwa untuk \\ n = 1, 2, 3, 4, \dots berlaku \\ \begin{aligned} (a + b)^{n} = & C_{0}^{n} a^{n} b^{0} + C_{1}^{n} a^{n - 1} b^{1} + C_{2}^{n} a^{n - 2} b^{2} \\ + C_{3}^{n} a^{n - 3} b^{3} + \dots + C_{n - 3}^{n} a^{3} b^{n - 3} \\ + C_{n - 2}^{n} a^{2} b^{n - 2} + C_{n - 1}^{n} a^{1} b^{n - 1} + C_{n}^{n} a^{0} b^{n} \\ = \sum_{r = 0}^{n} C_{r}^{n} a^{n - r} b^{r} \end{aligned} \end{aligned}$

$E. 2 Perluasan Binomial Newton$

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

$E. 3 Teorema Multinomial$

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

$CONTOH SOAL$

$\begin{array}{ll} 1. & Misalkan untuk n bilangan bulat \\ Positif. Tunjukklan bahwa \\ a . (1 + x)^{n} = \sum_{r = 0}^{n} C_{r}^{n} x^{r} = \sum_{r = 0}^{n} (\begin{array}{c} n \\ r \end{array}) x^{r} \\ b . (\begin{array}{c} n \\ 0 \end{array}) + (\begin{array}{c} n \\ 1 \end{array}) + (\begin{array}{c} n \\ 2 \end{array}) + \dots + (\begin{array}{c} n \\ n \end{array}) = 2^{n} \\ Bukti \\ \begin{aligned} a . (1 + x) & ^{n} \\ = & 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} + \dots + 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} \\ = & C_{0}^{n} + C_{1}^{n} x + C_{2}^{n} x^{2} + C_{3}^{n} x^{3} + \dots \\ + 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} \\ atau & dengan bentuk lain \\ = & (\begin{array}{c} n \\ 0 \end{array}) + (\begin{array}{c} n \\ 1 \end{array}) x + (\begin{array}{c} n \\ 2 \end{array}) x^{2} + (\begin{array}{c} n \\ 3 \end{array}) x^{3} \\ + \dots + (\begin{array}{c} n \\ n - 3 \end{array}) x^{n - 3} + (\begin{array}{c} n \\ n - 2 \end{array}) x^{n - 2} \\ + (\begin{array}{c} n \\ n - 1 \end{array}) x^{n - 1} + (\begin{array}{c} n \\ n \end{array}) x^{n} \\ = & \sum_{r = 0}^{n} (\begin{array}{c} n \\ r \end{array}) x^{r} \end{aligned} \\ \begin{aligned} b . (1 + x) & ^{n} lihat jawaban poin a, saat x = 1 \\ (1 + 1) & ^{n} = (\begin{array}{c} n \\ 0 \end{array}) + (\begin{array}{c} n \\ 1 \end{array}) 1 + (\begin{array}{c} n \\ 2 \end{array}) 1^{2} + (\begin{array}{c} n \\ 3 \end{array}) 1^{3} \\ + \dots + (\begin{array}{c} n \\ n - 3 \end{array}) 1^{n - 3} + (\begin{array}{c} n \\ n - 2 \end{array}) 1^{n - 2} \\ + (\begin{array}{c} n \\ n - 1 \end{array}) 1^{n - 1} + (\begin{array}{c} n \\ n \end{array}) 1^{n} \\ (2) & ^{n} = (\begin{array}{c} n \\ 0 \end{array}) + (\begin{array}{c} n \\ 1 \end{array}) + (\begin{array}{c} n \\ 2 \end{array}) + (\begin{array}{c} n \\ 3 \end{array}) \\ + \dots + (\begin{array}{c} n \\ n - 1 \end{array}) + (\begin{array}{c} n \\ n \end{array}) \\ = & \sum_{r = 0}^{n} (\begin{array}{c} n \\ r \end{array}) \\ Sehing & ga \\ 2^{n} & = \sum_{r = 0}^{n} (\begin{array}{c} n \\ r \end{array}) \end{aligned} \end{array}$

$\begin{array}{ll} 2. & Misalkan untuk n bilangan bulat \\ Positif. Tunjukklan bahwa \\ (\begin{array}{c} n \\ 0 \end{array}) - (\begin{array}{c} n \\ 1 \end{array}) + (\begin{array}{c} n \\ 2 \end{array}) - \dots + (- 1)^{n} (\begin{array}{c} n \\ n \end{array}) = 0 \\ Bukti \\ Sebelumnya diketahui bahwa \\ \begin{aligned} (a + b)^{n} = \sum_{r = 0}^{n} (\begin{array}{c} n \\ r \end{array}) a^{n - r} b^{r} \\ atau \\ \sum_{r = 0}^{n} (\begin{array}{c} n \\ r \end{array}) a^{n - r} b^{r} = (a + b)^{n} \\ ⧫ saat a = b = 1, maka \\ \sum_{r = 0}^{n} (\begin{array}{c} n \\ r \end{array}) 1^{n - r} 1^{r} = (1 + 1)^{n} \\ \Leftrightarrow \sum_{r = 0}^{n} (\begin{array}{c} n \\ r \end{array}) = 2^{n} . . . (bukti no. 1.b) \\ ⧫ saat a = 1 & b = - 1 maka \\ \sum_{r = 0}^{n} (\begin{array}{c} n \\ r \end{array}) 1^{n - r} (- 1)^{r} = (1 - 1)^{n} = 0 \\ Sehingga \\ (\begin{array}{c} n \\ 0 \end{array}) - (\begin{array}{c} n \\ 1 \end{array}) + (\begin{array}{c} n \\ 2 \end{array}) - \dots + (- 1)^{n} (\begin{array}{c} n \\ n \end{array}) = 0 ◼ \end{aligned} \end{array}$

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

$\begin{array}{ll} 4. & Tentukan nilai dari \\ a . (\begin{array}{c} 1 \\ 1 \end{array}) + (\begin{array}{c} 2 \\ 1 \end{array}) + (\begin{array}{c} 3 \\ 1 \end{array}) + \dots + (\begin{array}{c} 100 \\ 1 \end{array}) \\ b . (\begin{array}{c} 100 \\ 100 \end{array}) + (\begin{array}{c} 101 \\ 100 \end{array}) + (\begin{array}{c} 102 \\ 100 \end{array}) + \dots + (\begin{array}{c} 200 \\ 100 \end{array}) \\ Jawab : \\ \begin{aligned} a . & (\begin{array}{c} 1 \\ 1 \end{array}) + (\begin{array}{c} 2 \\ 1 \end{array}) + (\begin{array}{c} 3 \\ 1 \end{array}) + \dots + (\begin{array}{c} 100 \\ 1 \end{array}) \\ Sebelumnya perhatikan \\ = (\begin{array}{c} 1 \\ 1 \end{array}) + (\begin{array}{c} 2 \\ 1 \end{array}) + (\begin{array}{c} 3 \\ 1 \end{array}) + \dots + (\begin{array}{c} n \\ 1 \end{array}) \\ Karena \\ (\begin{array}{c} n \\ r - 1 \end{array}) + (\begin{array}{c} n \\ r \end{array}) = (\begin{array}{c} n + 1 \\ r \end{array}) \\ Saat \\ (\begin{array}{c} 1 \\ 1 \end{array}) + (\begin{array}{c} 2 \\ 1 \end{array}) = (\begin{array}{c} 2 \\ 2 \end{array}) + (\begin{array}{c} 2 \\ 1 \end{array}) = (\begin{array}{c} 3 \\ 2 \end{array}) \\ Sehingga \\ \underset{(\begin{array}{c} 5 \\ 2 \end{array})}{\underset{⏟}{\underset{(\begin{array}{c} 4 \\ 2 \end{array})}{\underset{⏟}{\underset{(\begin{array}{c} 3 \\ 2 \end{array})}{\underset{⏟}{(\begin{array}{c} 1 \\ 1 \end{array}) + (\begin{array}{c} 2 \\ 1 \end{array})}} + (\begin{array}{c} 3 \\ 1 \end{array})}} + (\begin{array}{c} 4 \\ 1 \end{array})}} \\ maka \\ = (\begin{array}{c} 1 \\ 1 \end{array}) + (\begin{array}{c} 2 \\ 1 \end{array}) + (\begin{array}{c} 3 \\ 1 \end{array}) + \dots + (\begin{array}{c} n \\ 1 \end{array}) = (\begin{array}{c} n + 1 \\ 2 \end{array}) \\ Jadi, \\ (\begin{array}{c} 1 \\ 1 \end{array}) + (\begin{array}{c} 2 \\ 1 \end{array}) + (\begin{array}{c} 3 \\ 1 \end{array}) + \dots + (\begin{array}{c} 100 \\ 1 \end{array}) = (\begin{array}{c} 101 \\ 2 \end{array}) \end{aligned} \\ b Silahkan coba sendiri \\ sebagai latihan \end{array}$

$\begin{array}{ll} 5. & Tentukanlah nilai dari \\ a . \sum_{r = 0}^{1000} (\begin{array}{c} 1000 \\ r \end{array}) \\ b . (\begin{array}{c} 2009 \\ 1 \end{array}) + (\begin{array}{c} 2009 \\ 2 \end{array}) + (\begin{array}{c} 2009 \\ 3 \end{array}) + \dots + (\begin{array}{c} 2009 \\ 2004 \end{array}) \\ (OSK 2009) \\ Jawab : \\ a . \sum_{r = 0}^{1000} (\begin{array}{c} 1000 \\ r \end{array}) = \sum_{r = 0}^{1000} (\begin{array}{c} 1000 \\ r \end{array}) 1^{1000 - r} 1^{r} \\ = (1 + 1)^{1000} = 2^{1000} \\ b . \sum_{r = 0}^{2009} (\begin{array}{c} 2009 \\ r \end{array}) = 2^{2009} \\ karena (\begin{array}{c} n \\ r \end{array}) = (\begin{array}{c} n \\ n - r \end{array}), maka \\ (\begin{array}{c} 2009 \\ 0 \end{array}) = (\begin{array}{c} 2009 \\ 2009 \end{array}), (\begin{array}{c} 2009 \\ 1 \end{array}) = (\begin{array}{c} 2009 \\ 2008 \end{array}), \\ \dots, (\begin{array}{c} 2009 \\ 1004 \end{array}) = (\begin{array}{c} 2009 \\ 1005 \end{array}) \\ Sehingga \\ (\begin{array}{c} 2009 \\ 0 \end{array}) + (\begin{array}{c} 2009 \\ 1 \end{array}) + (\begin{array}{c} 2009 \\ 2 \end{array}) + \dots + (\begin{array}{c} 2009 \\ 2009 \end{array}) = 2^{2009} \\ \Leftrightarrow (\begin{array}{c} 2009 \\ 0 \end{array}) + (\begin{array}{c} 2009 \\ 1 \end{array}) + (\begin{array}{c} 2009 \\ 2 \end{array}) + \dots + (\begin{array}{c} 2009 \\ 1004 \end{array}) = \frac{2^{2009}}{2} = 2^{2008} \\ \Leftrightarrow 1 + (\begin{array}{c} 2009 \\ 1 \end{array}) + (\begin{array}{c} 2009 \\ 2 \end{array}) + (\begin{array}{c} 2009 \\ 3 \end{array}) + \dots + (\begin{array}{c} 2009 \\ 1004 \end{array}) = 2^{2008} \\ \Leftrightarrow (\begin{array}{c} 2009 \\ 1 \end{array}) + (\begin{array}{c} 2009 \\ 2 \end{array}) + (\begin{array}{c} 2009 \\ 3 \end{array}) + \dots + (\begin{array}{c} 2009 \\ 1004 \end{array}) = 2^{2008} - 1 \end{array}$

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PAS MATEMATIKA BLOG

Binomial Newton

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