Deret Tak Hingga : Soal dan Pembahasan

Berikut ini adalah kumpulan soal mengenai deret tak hingga. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

1. \dfrac13+\dfrac2{3^2}+\dfrac3{3^3}+\dfrac4{3^4}+\cdots= ....
   
   Misal
   $\eqalign{
   s&=\dfrac13+\dfrac2{3^2}+\dfrac3{3^3}+\dfrac4{3^4}+\cdots\\
   \dfrac13s&=\dfrac1{3^2}+\dfrac2{3^3}+\dfrac3{3^4}+\dfrac4{3^5}+\cdots&\qquad-\\\hline
   \dfrac23s&=\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}+\cdots\\
   \dfrac23s&=\dfrac{\dfrac13}{1-\dfrac13}\\
   \dfrac23s&=\dfrac{\dfrac13}{\dfrac23}\\
   \dfrac23s&=\dfrac12\\
   s&=\dfrac12\cdot\dfrac32\\
   &=\boxed{\boxed{\dfrac34}}
   }$

   Jadi, \dfrac13+\dfrac2{3^2}+\dfrac3{3^3}+\dfrac4{3^4}+\cdots=\dfrac34.
   
   
2. 8+\dfrac{16}3+\dfrac83+\dfrac{32}{27}+\dfrac{40}{81}+\cdots=a\times\dfrac67. Maka nilai a adalah....
   
   1. 21
   2. 20

   3. 19
   4. 18
   Misal
   

   $\eqalign{
   s&=8+\dfrac{16}3+\dfrac83+\dfrac{32}{27}+\dfrac{40}{81}+\cdots\\
   s&=8+\dfrac{16}3+\dfrac{24}9+\dfrac{32}{27}+\dfrac{40}{81}+\cdots\\
   s&=8+\dfrac{16}3+\dfrac{24}{3^2}+\dfrac{32}{3^3}+\dfrac{40}{3^4}+\cdots\\
   \dfrac13s&=\dfrac83+\dfrac{16}{3^2}+\dfrac{24}{3^3}+\dfrac{32}{3^4}+\dfrac{40}{3^5}+\cdots&\qquad-\\\hline
   \dfrac23s&=8+\dfrac83+\dfrac8{3^2}+\dfrac8{3^3}+\dfrac8{3^4}+\cdots\\
   &=\dfrac8{1-\dfrac13}\\
   &=\dfrac8{\dfrac23}\\
   &=12\\
   s&=12\cdot\dfrac32\\
   a\times\dfrac67&=18\\
   a&=18\cdot\dfrac76\\
   &=\boxed{\boxed{21}}
   }$

   Jadi, nilai a adalah 21.
   JAWAB: A
   
   
3. 
   

j2