Soal dan Solusi #2

Pertanyaan 3
Koto Yunidar
mlm all, bntuin yak,
\frac{n!}{(n-3)!} = 210,  gmana caranya?
   
Jawaban 3
Ali Khan Su’ud


\frac{(n. (n-1).(n-2).(n-3)!)}{(n-3)!} = 210
n.(n-1).(n-2) = 210
itu bil brurutan.
7.6.5 =210
n = 7
Bagus Kepo Setiadi
210 itu kan 3.7.10 = 3.7.5.2
n.(n-1).(n-2) >> berarti berurutan kan
otak atik aja.. alhasil jadi 7.2.3.5 = 7.6.5
 


Pertanyaan 4
Putri Princezna
Tentukan hasil dari \dfrac{1}{2}+\dfrac{3}{4}+\dfrac{5}{8}+\dfrac{7}{16}+ \cdots
 
Jawaban 4
Hiyori Hinata
1+x+x^2+\cdots = \dfrac{1}{(1-x)} holds in (-1, 1). By using termwise differentiation, we get that
1+2x+3x^2+\cdots = \dfrac{1}{(1-x)^2}.
By substituting x with \frac{1}{2}, we have
\displaystyle\sum_{n=1}^{\infty}{\frac{n}{2^(n-1)}=4}
So
\begin{aligned}\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{5}{8}+\dfrac{7}{16}+ \cdots&=\displaystyle\sum_{n=1}^{\infty}{\frac{2n-1}{2^(n)}}\\&=\displaystyle \sum_{n=1}^{\infty}{\frac{n}{2^(n-1)}}-\sum_{n=1}^{\infty}{\frac{1}{2^(n)}}\\&=4-1= 3.\end{aligned}
 
Ashfaq Ahmad
\begin{aligned}S&= \frac{1}{2}+ \frac{3}{4}+ \frac{5}{8}+ \frac{7}{16}+\cdots\\ \frac{1}{2}S&= \frac{1}{4}+ \frac{3}{8}+ \frac{5}{16}+ \frac{7}{32}+\cdots\\ \text{So }&\\S-\frac{1}{2} S&= \frac{1}{2}+ \frac{2}{4}+ \frac{2}{8}+ \frac{2}{16}+\cdots\\ \frac{1}{2}S&= \frac{1}{2}+ \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+\cdots\\ \frac{1}{2}S&=\frac{1}{2}+1=\frac{3}{2}\\ \text{or }S&= 3\end{aligned}



Pertanyaan 5
Putri Princezna
Jika {}^2log \, a + {}^2log \, b = 12 dan 3 . {}^2log \, a- {}^2log \, b = 4,
maka a + b = \cdot
  
Jawaban 5
Purie Astagraha
{}^2log \, a= x
{}^2log \, b= y
{}^2log \, a + {}^2log \, b = 12 \to x+y=12
3 . {}^2log \, a- {}^2log \, b = 4 \to 3x-y=4
x= 4 \to a= 2^4
y= 8 \to b= 2^8
x+y= 2^4+2^8
 
Rusli Gustiono
hmmmm…………
{}^2log \, ab = 12 dan {}^2log \, \dfrac{a^3}{b} = 4
\dfrac{a^3}{b} = 2^4 \to a^3 =(2^4) b
ab=2^{12} \to a^3 . b^3 = 2^{36}
(2^4)b . b^3 = 2^{36} \to b^4 = 2^{32}
b = 2^8
a . 2^8 = 2^{12} \to a = 2^4
a + b =16 + 256
a + b = 272

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