HOTS Zone : Faktorial
Table of Contents
Tipe:
No.
Tentukan hasil dari \[\dfrac{(6!+5!)\times(4!+3!)\times(2!+1!)}{(6!-5!)\times(4!-3!)\times(2!-1!)}\]ALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{(6!+5!)\times(4!+3!)\times(2!+1!)}{(6!-5!)\times(4!-3!)\times(2!-1!)}&=\dfrac{5!(6+1)\times3!(4+1)\times1!(2+1)}{5!(6-1)\times3!(4-1)\times1!(2-1)}\\[3.5pt]
&=\dfrac{7\times5\times3}{5\times3\times1}\\
&=\boxed{\boxed{7}}
\end{aligned}
Jadi, \dfrac{(6!+5!)\times(4!+3!)\times(2!+1!)}{(6!-5!)\times(4!-3!)\times(2!-1!)}=7 .
No.
a! + b! + c! = d! dimana a, b, c, d adalah bilangan bulat positif. Berapa banyak solusiALTERNATIF PENYELESAIAN
d! ≥ 1! + 1! + 1! = 3
d ≥ 3
anggap a ≤ b ≤ c
\(\begin{aligned} d!&=a!+b!+c!\\ &\leq3c!\\ \dfrac{d!}{c!}&\leq3 \end{aligned}\)
yang memenuhi adalah d = 3 dan c = 2. Sehingga didapat a = 2 dan b = 2.
d ≥ 3
anggap a ≤ b ≤ c
\(\begin{aligned} d!&=a!+b!+c!\\ &\leq3c!\\ \dfrac{d!}{c!}&\leq3 \end{aligned}\)
yang memenuhi adalah d = 3 dan c = 2. Sehingga didapat a = 2 dan b = 2.
Jadi, hanya ada 1 solusi.
No.
Bentuk sederhana dari \[\dfrac{\sqrt3\left(n^2-1\right)!}{\left(n^2-3\right)!}\]ALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{\sqrt3\left(n^2-1\right)!}{\left(n^2-3\right)!}&=\dfrac{\sqrt3\left(n^2-1\right)\left(n^2-2\right)\left(n^2-3\right)!}{\left(n^2-3\right)!}\\[3.5pt]
&=\sqrt3\left(n^2-1\right)\left(n^2-2\right)\\
&=\boxed{\boxed{\sqrt3n^2-3\sqrt3n+2\sqrt3}}
\end{aligned}
Jadi, \dfrac{\sqrt3\left(n^2-1\right)!}{\left(n^2-3\right)!}=\sqrt3n^2-3\sqrt3n+2\sqrt3 .
No.
Diketahui \[\dfrac1{0!10!}+\dfrac1{1!9!}+\dfrac1{2!8!}+\cdots+\dfrac1{9!1!}+\dfrac1{10!0!}=\dfrac{m}n\] JikaALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac1{0!10!}+\dfrac1{1!9!}+\dfrac1{2!8!}+\cdots+\dfrac1{9!1!}+\dfrac1{10!0!}&=\dfrac{10!}{10!}\left(\dfrac1{0!10!}+\dfrac1{1!9!}+\dfrac1{2!8!}+\cdots+\dfrac1{9!1!}+\dfrac1{10!0!}\right)\\[3.8pt]
&=\dfrac1{10!}\left(\dfrac{10!}{0!10!}+\dfrac{10!}{1!9!}+\dfrac{10!}{2!8!}+\cdots+\dfrac{10!}{9!1!}+\dfrac{10!}{10!0!}\right)\\[3.8pt]
&=\dfrac1{10!}\left({10\choose0}+{10\choose1}+{10\choose2}+\cdots+{10\choose9}+{10\choose10}\right)\\[3.8pt]
&=\dfrac1{10!}\left(2^{10}\right)\\[3.8pt]
&=\dfrac{\cancel{2}\cdot\cancel{2\cdot2\cdot2}\cdot\cancel{2}\cdot\cancel{2\cdot2}\cdot\cancel2\cdot2\cdot2}{\cancelto{5}{10}\cdot9\cdot\cancel{8}\cdot7\cdot\cancelto{3}6\cdot5\cdot\cancel4\cdot3\cdot\cancel2}\\[3.8pt]
&=\dfrac4{14175}
\end{aligned}\)
10(4) = 40
10(4) = 40
Jadi, nilai dari 10m adalah 40.
No.
Bentuk sederhana dari :\[2\cdot1!+4\cdot2!+6\cdot3!+\cdots+200\cdot100!\] adalah ....- 100! − 2
- 101! − 2
- 2(100! − 1)
- 2(101! − 1)
- 2(101! + 1)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
2\cdot1!+4\cdot2!+6\cdot3!+\cdots+200\cdot100!&=2\left(1\cdot1!+2\cdot2!+3\cdot3!+\cdots+100\cdot100!\right)\\
&=2\left((2-1)\cdot1!+(3-1)\cdot2!+(4-1)\cdot3!+\cdots+(101-1)\cdot100!\right)\\
&=2\left(2\cdot1!-1!+3\cdot2!-2!+4\cdot3!-3!+\cdots+101\cdot100!-100!\right)\\
&=2\left(2!-1!+3!-2!+4!-3!+\cdots+101!-100!\right)\\
&=2\left(-1!+101!\right)\\
&=\color{blue}\boxed{\boxed{\color{black}2\left(101!-1\right)}}
\end{aligned}\)
Jadi, bentuk sederhana dari :\[2\cdot1!+4\cdot2!+6\cdot3!+\cdots+200\cdot100!\]
adalah 2(101! − 1).
JAWAB: C
JAWAB: C
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