Soal dan Solusi #2

Pertanyaan 3
Koto Yunidar
mlm all, bntuin yak,

$\frac{n!}{(n-3)!} = 210,$ gmana caranya?

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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 ${}^2log \, \dfrac{a^3}{b} = 4$
$\dfrac{a^3}{b} = 2^4 \to a^3 =(2^4) b$
$ab=2^{12} \to a^3 . b^3 = 2^{36}$
$(2^4)b . b^3 = 2^{36} \to b^4 = 2^{32}$
$b = 2^8$
$a . 2^8 = 2^{12} \to a = 2^4$
$a + b =16 + 256$
$a + b = 272$

Soal dan Solusi #2

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