Beranda > soal math > Soal dan Solusi #2

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,

\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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