Dengan
Jabarkan (x+1)^3
Jabarkan(x-2)^4
\(\eqalign{
(x+1)^3&=\displaystyle\sum_{k=0}^3{3\choose k}x^{3-k}1^k\\
&=\displaystyle\sum_{k=0}^3{3\choose k}x^{3-k}\\
&={3\choose 0}x^3+{3\choose 1}x^2+{3\choose 2}x+{3\choose 3}x^0\\
&=1x^3+3x^2+3x+1(1)\\
&=x^3+3x^2+3x+1
}\)
Jadi, (x+1)^3=x^3+3x^2+3x+1
Jabarkan
\(\eqalign{
(x-2)^4&=\displaystyle\sum_{k=0}^4{4\choose k}x^{4-k}(-2)^k\\
&={4\choose 0}x^{4-0}(-2)^0+{4\choose 1}x^{4-1}(-2)^1+{4\choose 2}x^{4-2}(-2)^2+{4\choose 3}x^{4-3}(-2)^3+{4\choose 4}x^{4-4}(-2)^4\\
&=1x^4(1)+4x^3(-2)+6x^2(4)+4x^1(-8)+1x^0(16)\\
&=x^4-8x^3+24x^2-32x+16
}\)
Jadi, (x-2)^4=x^4-8x^3+24x^2-32x+16
KOEFISIEN PADA SUKU TERTENTU
Koefisien pada suku ke-
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}
\(\eqalign{ x^{7-k}&=x^5\\ 7-k&=5\\ k&=2 }\)
\(\eqalign{ {7\choose k}4^k&={7\choose 2}4^2\\ &=\dfrac{7!}{(7-2)!\cdot2!}(16)\\ &=\dfrac{7\cdot6\cdot5!}{5!\cdot2\cdot1}(16)\\ &=(21)(16)\\ &=\boxed{\boxed{336}} }\)
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
\(\eqalign{ {4\choose k}(-5)^k&={4\choose 2}(-5)^2\\ &=\dfrac{4!}{(4-2)!\cdot2!}(25)\\ &=\dfrac{4\cdot3\cdot2!}{2!\cdot2\cdot1}(25)\\ &=(6)(25)\\ &=\boxed{\boxed{150}} }\)
\(\eqalign{ x^{7-k}&=x^5\\ 7-k&=5\\ k&=2 }\)
\(\eqalign{ {7\choose k}4^k&={7\choose 2}4^2\\ &=\dfrac{7!}{(7-2)!\cdot2!}(16)\\ &=\dfrac{7\cdot6\cdot5!}{5!\cdot2\cdot1}(16)\\ &=(21)(16)\\ &=\boxed{\boxed{336}} }\)
Jadi, koefisien x^5 dari penjabaran (x+4)^7 adalah 336 .
Carilah koefisien suku ke-
\(\eqalign{ {4\choose k}(-5)^k&={4\choose 2}(-5)^2\\ &=\dfrac{4!}{(4-2)!\cdot2!}(25)\\ &=\dfrac{4\cdot3\cdot2!}{2!\cdot2\cdot1}(25)\\ &=(6)(25)\\ &=\boxed{\boxed{150}} }\)
Jadi, koefisien suku ke-3 dari penjabaran (x-5)^4 adalah 150 .
RUMUS-RUMUS KHUSUS
Dengan
\(\eqalign{
\displaystyle\sum_{k=0}^n{n\choose k}x^k&=\displaystyle\sum_{k=0}^n{n\choose k}1^{n-k}x^k\\
&=(1+x)^n
}\)
J
Dengan
\(\eqalign{
\displaystyle\sum_{k=0}^n{n\choose k}&=\displaystyle\sum_{k=0}^n{n\choose k}1^k\\
&=(1+1)^n\\
&=2^n
}\)
J
Dengan
Kita turunkan,
Untuk
\(\eqalign{ \displaystyle\sum_{k=0}^nk{n\choose k}1^{k-1}&=n(1+1)^{n-1}\\ \displaystyle\sum_{k=0}^nk{n\choose k}&=n2^{n-1} }\)
J
Dengan
Kita turunkan,
Untuk
\(\eqalign{ \displaystyle\sum_{k=0}^n\left(k^2-k\right){n\choose k}1^{k-2}&=\left(n^2-n\right)(1+1)^{n-2}\\ \displaystyle\sum_{k=0}^n\left(k^2-k\right){n\choose k}&=\left(n^2-n\right)2^{n-2}\\ \displaystyle\sum_{k=0}^nk^2{n\choose k}-\displaystyle\sum_{k=0}^nk{n\choose k}&=\left(n^2-n\right)2^{n-2}\\ \displaystyle\sum_{k=0}^nk^2{n\choose k}-n2^{n-1}&=\left(n^2-n\right)2^{n-2}\\ \displaystyle\sum_{k=0}^nk^2{n\choose k}&=\left(n^2-n\right)2^{n-2}+n2^{n-1}\\ &=\left(n^2-n\right)2^{n-2}+2n2^{n-2}\\ &=\left(n^2-n+2n\right)2^{n-2}\\ &=\left(n^2+n\right)2^{n-2} }\)
J
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