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

\(\eqalign{ (x+1)^3&=\displaystyle\sum_{k=0}^3{3\choose k}x^{3-k}1^k\\ &=\displaystyle\sum_{k=0}^3{3\choose k}x^{3-k}\\ &={3\choose 0}x^3+{3\choose 1}x^2+{3\choose 2}x+{3\choose 3}x^0\\ &=1x^3+3x^2+3x+1(1)\\ &=x^3+3x^2+3x+1 }\)

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

Jabarkan (x-2)^4

\(\eqalign{ (x-2)^4&=\displaystyle\sum_{k=0}^4{4\choose k}x^{4-k}(-2)^k\\ &={4\choose 0}x^{4-0}(-2)^0+{4\choose 1}x^{4-1}(-2)^1+{4\choose 2}x^{4-2}(-2)^2+{4\choose 3}x^{4-3}(-2)^3+{4\choose 4}x^{4-4}(-2)^4\\ &=1x^4(1)+4x^3(-2)+6x^2(4)+4x^1(-8)+1x^0(16)\\ &=x^4-8x^3+24x^2-32x+16 }\)

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}

\(\eqalign{ x^{7-k}&=x^5\\ 7-k&=5\\ k&=2 }\)

\(\eqalign{ {7\choose k}4^k&={7\choose 2}4^2\\ &=\dfrac{7!}{(7-2)!\cdot2!}(16)\\ &=\dfrac{7\cdot6\cdot5!}{5!\cdot2\cdot1}(16)\\ &=(21)(16)\\ &=\boxed{\boxed{336}} }\)

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

\(\eqalign{ {4\choose k}(-5)^k&={4\choose 2}(-5)^2\\ &=\dfrac{4!}{(4-2)!\cdot2!}(25)\\ &=\dfrac{4\cdot3\cdot2!}{2!\cdot2\cdot1}(25)\\ &=(6)(25)\\ &=\boxed{\boxed{150}} }\)

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

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

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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

\(\eqalign{ \displaystyle\sum_{k=0}^n{n\choose k}&=\displaystyle\sum_{k=0}^n{n\choose k}1^k\\ &=(1+1)^n\\ &=2^n }\)

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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,
\(\eqalign{ \displaystyle\sum_{k=0}^nk{n\choose k}1^{k-1}&=n(1+1)^{n-1}\\ \displaystyle\sum_{k=0}^nk{n\choose k}&=n2^{n-1} }\)

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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,
\(\eqalign{ \displaystyle\sum_{k=0}^n\left(k^2-k\right){n\choose k}1^{k-2}&=\left(n^2-n\right)(1+1)^{n-2}\\ \displaystyle\sum_{k=0}^n\left(k^2-k\right){n\choose k}&=\left(n^2-n\right)2^{n-2}\\ \displaystyle\sum_{k=0}^nk^2{n\choose k}-\displaystyle\sum_{k=0}^nk{n\choose k}&=\left(n^2-n\right)2^{n-2}\\ \displaystyle\sum_{k=0}^nk^2{n\choose k}-n2^{n-1}&=\left(n^2-n\right)2^{n-2}\\ \displaystyle\sum_{k=0}^nk^2{n\choose k}&=\left(n^2-n\right)2^{n-2}+n2^{n-1}\\ &=\left(n^2-n\right)2^{n-2}+2n2^{n-2}\\ &=\left(n^2-n+2n\right)2^{n-2}\\ &=\left(n^2+n\right)2^{n-2} }\)

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