Soal dan Solusi #2

Diambil dari grup facebook soul-mate-matika, ketika dulu saya jadi adminnya bro. :p Kemudian saya buatkan arsipnya di blog soul-mate-matika yang saya buat juga. https://soulmatematika.wordpress.com/category/soul-mate-matika/

Sekarang saya posting ulang di blog ini supaya jadi satu kesatuan, yaitu asimtot, membahas masalah matematika. :p

Pertanyaan 3

Koto Yunidar

mlm all, bntuin yak,

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

   

Jawaban 3

Ali Khan Su'ud

$latex \frac{(n. (n-1).(n-2).(n-3)!)}{(n-3)!} = 210$

$latex n.(n-1).(n-2) = 210$

itu bil brurutan.

$latex 7.6.5 =210$

$latex n = 7$

Bagus Kepo Setiadi

$latex 210$ itu kan $latex 3.7.10 = 3.7.5.2$
$latex n.(n-1).(n-2)$ >> berarti berurutan kan
otak atik aja.. alhasil jadi $latex 7.2.3.5 = 7.6.5$

 

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Pertanyaan 4

Putri Princezna

Tentukan hasil dari $latex \dfrac{1}{2}+\dfrac{3}{4}+\dfrac{5}{8}+\dfrac{7}{16}+ \cdots$

 

Jawaban 4

Hiyori Hinata

$latex 1+x+x^2+\cdots = \dfrac{1}{(1-x)}$ holds in $latex (-1, 1).$ By using termwise differentiation, we get that

$latex 1+2x+3x^2+\cdots = \dfrac{1}{(1-x)^2}.$

By substituting $latex x$ with $latex \frac{1}{2},$ we have

$latex \displaystyle\sum_{n=1}^{\infty}{\frac{n}{2^(n-1)}=4}$

So

$latex \begin{aligned}\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{5}{8}+\dfrac{7}{16}+ \cdots&=\displaystyle\sum_{n=1}^{\infty}{\frac{2n-1}{2^(n)}}\\&=\displaystyle \sum_{n=1}^{\infty}{\frac{n}{2^(n-1)}}-\sum_{n=1}^{\infty}{\frac{1}{2^(n)}}\\&=4-1= 3.\end{aligned}$

 

Ashfaq Ahmad

$latex \begin{aligned}S&= \frac{1}{2}+ \frac{3}{4}+ \frac{5}{8}+ \frac{7}{16}+\cdots\\ \frac{1}{2}S&= \frac{1}{4}+ \frac{3}{8}+ \frac{5}{16}+ \frac{7}{32}+\cdots\\ \text{So }&\\S-\frac{1}{2} S&= \frac{1}{2}+ \frac{2}{4}+ \frac{2}{8}+ \frac{2}{16}+\cdots\\ \frac{1}{2}S&= \frac{1}{2}+ \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+\cdots\\ \frac{1}{2}S&=\frac{1}{2}+1=\frac{3}{2}\\ \text{or }S&= 3\end{aligned}$

 

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Pertanyaan 5

Putri Princezna

Jika $latex {}^2log \, a + {}^2log \, b = 12$ dan $latex 3 . {}^2log \, a- {}^2log \, b = 4$,
maka $latex a + b = \cdot$

  

Jawaban 5

Purie Astagraha

$latex {}^2log \, a= x$

$latex {}^2log \, b= y$

$latex {}^2log \, a + {}^2log \, b = 12 \to x+y=12$

$latex 3 . {}^2log \, a- {}^2log \, b = 4 \to 3x-y=4$

$latex x= 4 \to a= 2^4$

$latex y= 8 \to b= 2^8$

$latex x+y= 2^4+2^8$

 

Rusli Gustiono

hmmmm............

$latex {}^2log \, ab = 12$ dan $latex {}^2log \, \dfrac{a^3}{b} = 4$

$latex \dfrac{a^3}{b} = 2^4 \to a^3 =(2^4) b$

$latex ab=2^{12} \to a^3 . b^3 = 2^{36}$

$latex (2^4)b . b^3 = 2^{36} \to b^4 = 2^{32}$

$latex b = 2^8$

$latex a . 2^8 = 2^{12} \to a = 2^4$

$latex a + b =16 + 256$

$latex a + b = 272$

 

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