Koefisien Binomial

Untuk menjabarkan (x+y)^n kita gunakan rumus:

(x+y)^n=\displaystyle\sum_{k=0}^n{n\choose k}x^{n-k}y^k

\displaystyle{n\choose k}=\dfrac{n!}{(n-k)!\cdot k!}
Dengan k bilangan cacah

Jabarkan (x+1)^3

$\begin{array}{rl}(x+1{)}^3& ={\displaystyle \sum _{k=0}^3(\genfrac{}{}{0ex}{}{3}{k})x^{3-k}{1}^k}\\ & ={\displaystyle \sum _{k=0}^3(\genfrac{}{}{0ex}{}{3}{k})x^{3-k}}\\ & =(\genfrac{}{}{0ex}{}{3}{0})x^3+(\genfrac{}{}{0ex}{}{3}{1})x^2+(\genfrac{}{}{0ex}{}{3}{2})x+(\genfrac{}{}{0ex}{}{3}{3})x^0\\ & =1x^3+3x^2+3x+1(1)\\ & =x^3+3x^2+3x+1\end{array}$

Jadi, (x+1)^3=x^3+3x^2+3x+1

Jabarkan (x-2)^4

$\begin{array}{rl}(x-2{)}^4& ={\displaystyle \sum _{k=0}^4(\genfrac{}{}{0ex}{}{4}{k})x^{4-k}(-2{)}^k}\\ & =(\genfrac{}{}{0ex}{}{4}{0})x^{4-0}(-2{)}^0+(\genfrac{}{}{0ex}{}{4}{1})x^{4-1}(-2{)}^1+(\genfrac{}{}{0ex}{}{4}{2})x^{4-2}(-2{)}^2+(\genfrac{}{}{0ex}{}{4}{3})x^{4-3}(-2{)}^3+(\genfrac{}{}{0ex}{}{4}{4})x^{4-4}(-2{)}^4\\ & =1x^4(1)+4x^3(-2)+6x^2(4)+4x^1(-8)+1x^0(16)\\ & =x^4-8x^3+24x^2-32x+16\end{array}$

Jadi, (x-2)^4=x^4-8x^3+24x^2-32x+16

KOEFISIEN PADA SUKU TERTENTU

Koefisien pada suku ke-p maka k=p-1

Tentukan koefisien x^5 dari penjabaran (x+4)^7

(x+4)^7=\displaystyle\sum_{k=0}^7{7\choose k}x^{7-k}4^k=\displaystyle\sum_{k=0}^7{7\choose k}4^kx^{7-k}

$\begin{array}{rl}{x}^{7-k}& =x^5\\ 7-k& =5\\ k& =2\end{array}$

$\begin{array}{rl}(\genfrac{}{}{0ex}{}{7}{k})4^k& =(\genfrac{}{}{0ex}{}{7}{2})4^2\\ & ={\displaystyle \frac{7!}{(7-2)!\cdot 2!}}(16)\\ & ={\displaystyle \frac{7\cdot 6\cdot 5!}{5!\cdot 2\cdot 1}}(16)\\ & =(21)(16)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 336}}}}\end{array}$

Jadi, koefisien x^5 dari penjabaran (x+4)^7 adalah 336.

Carilah koefisien suku ke-3 dari penjabaran (x-5)^4

(x-5)^4=\displaystyle\sum_{k=0}^4{4\choose k}x^{4-k}(-5)^k=\displaystyle\sum_{k=0}^4{4\choose k}(-5)^kx^{4-k}

k=3-1=2

$\begin{array}{rl}(\genfrac{}{}{0ex}{}{4}{k})(-5{)}^k& =(\genfrac{}{}{0ex}{}{4}{2})(-5{)}^2\\ & ={\displaystyle \frac{4!}{(4-2)!\cdot 2!}}(25)\\ & ={\displaystyle \frac{4\cdot 3\cdot 2!}{2!\cdot 2\cdot 1}}(25)\\ & =(6)(25)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 150}}}}\end{array}$

Jadi, koefisien suku ke-3 dari penjabaran (x-5)^4 adalah 150.

RUMUS-RUMUS KHUSUS

\displaystyle\sum_{k=0}^n{n\choose k}x^k=(1+x)^n

\displaystyle{n\choose k}=\dfrac{n!}{(n-k)!\cdot k!}
Dengan k bilangan cacah

$\begin{array}{rl}{\displaystyle \sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^k}& ={\displaystyle \sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})1^{n-k}{x}^k}\\ & =(1+x{)}^n\end{array}$

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\displaystyle\sum_{k=0}^n{n\choose k}=2^n

\displaystyle{n\choose k}=\dfrac{n!}{(n-k)!\cdot k!}
Dengan k bilangan cacah

$\begin{array}{rl}{\displaystyle \sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})}& ={\displaystyle \sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})1^k}\\ & =(1+1{)}^n\\ & =2^n\end{array}$

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\displaystyle\sum_{k=0}^nk{n\choose k}=n2^{n-1}

\displaystyle{n\choose k}=\dfrac{n!}{(n-k)!\cdot k!}
Dengan k bilangan cacah

\displaystyle\sum_{k=0}^n{n\choose k}x^k=(1+x)^n
Kita turunkan, \displaystyle\sum_{k=0}^nk{n\choose k}x^{k-1}=n(1+x)^{n-1}
Untuk x=1,
$\begin{array}{rl}{\displaystyle \sum _{k=0}^{n}k(\genfrac{}{}{0ex}{}{n}{k})1^{k-1}}& =n(1+1{)}^{n-1}\\ {\displaystyle \sum _{k=0}^{n}k(\genfrac{}{}{0ex}{}{n}{k})}& =n{2}^{n-1}\end{array}$

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\displaystyle\sum_{k=0}^nk^2{n\choose k}=\left(n^2+n\right)2^{n-2}

\displaystyle{n\choose k}=\dfrac{n!}{(n-k)!\cdot k!}
Dengan k bilangan cacah

\displaystyle\sum_{k=0}^nk{n\choose k}x^{k-1}=n(1+x)^{n-1}
Kita turunkan, \displaystyle\sum_{k=0}^n\left(k^2-k\right){n\choose k}x^{k-2}=\left(n^2-n\right)(1+x)^{n-2}
Untuk x=1,
$\begin{array}{rl}{\displaystyle \sum _{k=0}^n(k^2-k)(\genfrac{}{}{0ex}{}{n}{k})1^{k-2}}& =(n^2-n)(1+1{)}^{n-2}\\ {\displaystyle \sum _{k=0}^n(k^2-k)(\genfrac{}{}{0ex}{}{n}{k})}& =(n^2-n)2^{n-2}\\ {\displaystyle \sum _{k=0}^n{k}^2(\genfrac{}{}{0ex}{}{n}{k})-{\displaystyle \sum _{k=0}^{n}k(\genfrac{}{}{0ex}{}{n}{k})}}& =(n^2-n)2^{n-2}\\ {\displaystyle \sum _{k=0}^n{k}^2(\genfrac{}{}{0ex}{}{n}{k})-n{2}^{n-1}}& =(n^2-n)2^{n-2}\\ {\displaystyle \sum _{k=0}^n{k}^2(\genfrac{}{}{0ex}{}{n}{k})}& =(n^2-n)2^{n-2}+n{2}^{n-1}\\ & =(n^2-n)2^{n-2}+2n{2}^{n-2}\\ & =(n^2-n+2n)2^{n-2}\\ & =(n^2+n)2^{n-2}\end{array}$

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